Scanner is skipping nextLine() after using next() or nextFoo()?

537

I am using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    

int option;

option = input.nextInt(); // Read numerical value from input

System.out.println("Enter 1st string"); 

String string1 = input.nextLine(); // Read 1st string (this is skipped)

System.out.println("Enter 2nd string");

String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:

Enter numerical value

3   // This is my input

Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped

Enter 2nd string    // ...and this line is executed and waits for my input

I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.

edited Apr 4 '18 at 13:31

Lii

6,88544159

asked Oct 27 '12 at 16:37

blekione

3,00131016

4

Related: stackoverflow.com/questions/4708219/…

– James P.
Aug 14 '11 at 12:27

25

The newline character is probably not consumed.

– Lews Therin
Oct 27 '12 at 16:39

5

Style note: you should declare your String variables on the same line you assign a value. E.g. String string1 = input.nextLine ().

– Duncan Jones
Oct 27 '12 at 16:39

2

@Lews Therin: Yes, you are right. After another couple of test I found that it is executed and read "n". How to avoid of that?

– blekione
Oct 27 '12 at 16:45

3

Or you could be like me and use BufferedReader :) I don't care if it's old school, it has always worked and always will work for me. Also, knowledge of BufferedReader has application elsewhere. I simply don't like Scanner.

– Cruncher
Aug 27 '13 at 19:36

add a comment |

537

I am using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    

int option;

option = input.nextInt(); // Read numerical value from input

System.out.println("Enter 1st string"); 

String string1 = input.nextLine(); // Read 1st string (this is skipped)

System.out.println("Enter 2nd string");

String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:

Enter numerical value

3   // This is my input

Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped

Enter 2nd string    // ...and this line is executed and waits for my input

edited Apr 4 '18 at 13:31

Lii

6,88544159

asked Oct 27 '12 at 16:37

blekione

3,00131016

4

Related: stackoverflow.com/questions/4708219/…

– James P.
Aug 14 '11 at 12:27

25

The newline character is probably not consumed.

– Lews Therin
Oct 27 '12 at 16:39

5

Style note: you should declare your String variables on the same line you assign a value. E.g. String string1 = input.nextLine ().

– Duncan Jones
Oct 27 '12 at 16:39

2

@Lews Therin: Yes, you are right. After another couple of test I found that it is executed and read "n". How to avoid of that?

– blekione
Oct 27 '12 at 16:45

3

Or you could be like me and use BufferedReader :) I don't care if it's old school, it has always worked and always will work for me. Also, knowledge of BufferedReader has application elsewhere. I simply don't like Scanner.

– Cruncher
Aug 27 '13 at 19:36

add a comment |

537

245

I am using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    

int option;

option = input.nextInt(); // Read numerical value from input

System.out.println("Enter 1st string"); 

String string1 = input.nextLine(); // Read 1st string (this is skipped)

System.out.println("Enter 2nd string");

String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:

Enter numerical value

3   // This is my input

Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped

Enter 2nd string    // ...and this line is executed and waits for my input

edited Apr 4 '18 at 13:31

Lii

6,88544159

asked Oct 27 '12 at 16:37

blekione

3,00131016

I am using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    

int option;

option = input.nextInt(); // Read numerical value from input

System.out.println("Enter 1st string"); 

String string1 = input.nextLine(); // Read 1st string (this is skipped)

System.out.println("Enter 2nd string");

String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:

Enter numerical value

3   // This is my input

Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped

Enter 2nd string    // ...and this line is executed and waits for my input

java io java.util.scanner

edited Apr 4 '18 at 13:31

Lii

6,88544159

asked Oct 27 '12 at 16:37

blekione

3,00131016

edited Apr 4 '18 at 13:31

Lii

6,88544159

asked Oct 27 '12 at 16:37

blekione

3,00131016

edited Apr 4 '18 at 13:31

Lii

6,88544159

edited Apr 4 '18 at 13:31

Lii

6,88544159

edited Apr 4 '18 at 13:31

Lii

6,88544159

asked Oct 27 '12 at 16:37

blekione

3,00131016

asked Oct 27 '12 at 16:37

blekione

3,00131016

asked Oct 27 '12 at 16:37

blekione

3,00131016

4

Related: stackoverflow.com/questions/4708219/…

– James P.
Aug 14 '11 at 12:27

25

The newline character is probably not consumed.

– Lews Therin
Oct 27 '12 at 16:39

5

Style note: you should declare your String variables on the same line you assign a value. E.g. String string1 = input.nextLine ().

– Duncan Jones
Oct 27 '12 at 16:39

2

@Lews Therin: Yes, you are right. After another couple of test I found that it is executed and read "n". How to avoid of that?

– blekione
Oct 27 '12 at 16:45

3

Or you could be like me and use BufferedReader :) I don't care if it's old school, it has always worked and always will work for me. Also, knowledge of BufferedReader has application elsewhere. I simply don't like Scanner.

– Cruncher
Aug 27 '13 at 19:36

add a comment |

4

Related: stackoverflow.com/questions/4708219/…

– James P.
Aug 14 '11 at 12:27

25

The newline character is probably not consumed.

– Lews Therin
Oct 27 '12 at 16:39

5

Style note: you should declare your String variables on the same line you assign a value. E.g. String string1 = input.nextLine ().

– Duncan Jones
Oct 27 '12 at 16:39

2

@Lews Therin: Yes, you are right. After another couple of test I found that it is executed and read "n". How to avoid of that?

– blekione
Oct 27 '12 at 16:45

3

Or you could be like me and use BufferedReader :) I don't care if it's old school, it has always worked and always will work for me. Also, knowledge of BufferedReader has application elsewhere. I simply don't like Scanner.

– Cruncher
Aug 27 '13 at 19:36

Related: stackoverflow.com/questions/4708219/…

– James P.
Aug 14 '11 at 12:27

The newline character is probably not consumed.

– Lews Therin
Oct 27 '12 at 16:39

Style note: you should declare your String variables on the same line you assign a value. E.g. String string1 = input.nextLine ().

– Duncan Jones
Oct 27 '12 at 16:39

@Lews Therin: Yes, you are right. After another couple of test I found that it is executed and read "n". How to avoid of that?

– blekione
Oct 27 '12 at 16:45

Or you could be like me and use BufferedReader :) I don't care if it's old school, it has always worked and always will work for me. Also, knowledge of BufferedReader has application elsewhere. I simply don't like Scanner.

– Cruncher
Aug 27 '13 at 19:36

add a comment |

15 Answers
15

active

oldest

votes

634

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
```
int option = input.nextInt();

input.nextLine();  // Consume newline left-over

String str1 = input.nextLine();
```

Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
```
int option = 0;

try {

    option = Integer.parseInt(input.nextLine());

} catch (NumberFormatException e) {

    e.printStackTrace();

}

String str1 = input.nextLine();
```

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above?

– blekione
Oct 27 '12 at 17:52

3

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right?

– Rohit Jain
Oct 27 '12 at 18:02

3

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions.

– Rohit Jain
Oct 27 '12 at 18:02

4

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign.

– Mordechai
Jan 11 '17 at 3:00

1

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too.

– MildlyMilquetoast
Jan 22 '17 at 1:13

|
show 4 more comments

179

The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.

Try it like that:

System.out.print("Insert a number: ");

int number = input.nextInt();

input.nextLine(); // This line you have to add (It consumes the n character)

System.out.print("Text1: ");

String text1 = input.nextLine();

System.out.print("Text2: ");

String text2 = input.nextLine();

edited Dec 29 '17 at 6:57

Ebony Maw

316314

answered Aug 14 '11 at 12:24

Prine

8,59773355

umm seems a solution, making the skipped line non-used nextLine, but I still need an explanation of this behaviour

– Eng.Fouad
Aug 14 '11 at 12:25

2

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:11

add a comment |

It's because when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line". When input.nextLine() executes, it consumes the "end of line" still in the buffer from the first input.

Instead, use input.nextLine() immediately after input.nextInt()

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

Mmm, that bug was corrected in now a days?

– Victor
May 29 '14 at 13:28

7

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input.

– Bohemian♦
May 29 '14 at 13:38

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP.

– Victor
May 29 '14 at 14:55

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page.

– Bohemian♦
May 29 '14 at 15:18

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[rn]+") to drop any newline characters after calling nextInt().

EDIT: as @PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip: https://stackoverflow.com/a/42471816/143585

edited May 23 '17 at 12:10

Community♦

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

I know what we do to remove the data in buffer, but this case, please help me with this: stackoverflow.com/questions/33585314/having-issues-with-scanner

– Khuong
Nov 7 '15 at 18:03

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: stackoverflow.com/a/42471816/7098259

– Patrick Parker
Feb 26 '17 at 17:52

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours.

– Denis Tulskiy
Feb 27 '17 at 5:57

@PatrickParker Why does is cause infinite loop ?

– Piyush
Aug 19 '17 at 8:41

|
show 1 more comment

It does that because input.nextInt(); doesn't capture the newline. you could do like the others proposed by adding an input.nextLine(); underneath.

Alternatively you can do it C# style and parse a nextLine to an integer like so:

int number = Integer.parseInt(input.nextLine());

Doing this works just as well, and it saves you a line of code.

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Things you need to know:

text which represents few lines also contains non-printable characters between lines (we call them line separators) like
- carriage return (CR - in String literals represented as "r")
- line feed (LF - in String literals represented as "n")

when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows rn) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42rn".

Problem

Scanner#nextInt (and other Scanner#nextType methods) doesn't allow Scanner to consume these line separators. It will read them from System.in (how else Scanner would know that there are no more digits from the user which represent age value than facing whitespace?) which will remove them from standard input, but it will also cache those line separators internally. What we need to remember, is that all of the Scanner methods are always scanning starting from the cached text.

Now Scanner#nextLine() simply collects and returns all characters until it finds line separators (or end of stream). But since line separators after reading the number from the console are found immediately in Scanner's cache, it returns empty String, meaning that Scanner was not able to find any character before those line separators (or end of stream).

BTW nextLine also consumes those line separators.

Solution

So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either

consume line separator left by nextInt from Scanners cache by
- calling nextLine,
- or by calling skip("\R") to let Scanner skip part matched by R which represents line separator (more info about R: https://stackoverflow.com/a/31060125)

don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.

BTW: Scanner#nextType methods can skip delimiters (by default all whitespaces like tabs, line separators) including those cached by scanner, until they will find next non-delimiter value (token). Thanks to that for input like "42rnrn321rnrnrnfoobar" code

int num1 = sc.nextInt();

int num2 = sc.nextInt();

String name = sc.next();

will be able to properly assign num1=42 num2=321 name=foobar.

edited Aug 15 '18 at 16:29

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

add a comment |

Instead of input.nextLine() use input.next(), that should solve the problem.

Modified code:

public static Scanner input = new Scanner(System.in);



public static void main(String args)

{

    System.out.print("Insert a number: ");

    int number = input.nextInt();

    System.out.print("Text1: ");

    String text1 = input.next();

    System.out.print("Text2: ");

    String text2 = input.next();

}

edited Jul 23 '14 at 10:31

arghtype

3,04192846

answered Jul 23 '14 at 10:20

Castaldi

374417

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

In order to avoid the issue, use nextLine(); immediately after nextInt(); as it helps in clearing out the buffer. When you press ENTER the nextInt(); does not capture the new line and hence, skips the Scanner code later.

Scanner scanner =  new Scanner(System.in);

int option = scanner.nextInt();

scanner.nextLine(); //clearing the buffer

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

add a comment |

If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .

Code :

class ScanReader {

/**

* @author Nikunj Khokhar

*/

    private byte buf = new byte[4 * 1024];

    private int index;

    private BufferedInputStream in;

    private int total;



    public ScanReader(InputStream inputStream) {

        in = new BufferedInputStream(inputStream);

    }



    private int scan() throws IOException {

        if (index >= total) {

            index = 0;

            total = in.read(buf);

            if (total <= 0) return -1;

        }

        return buf[index++];

    }

    public char scanChar(){

        int c=scan();

        while (isWhiteSpace(c))c=scan();

        return (char)c;

    }





    public int scanInt() throws IOException {

        int integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public String scanString() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        StringBuilder res = new StringBuilder();

        do {

            res.appendCodePoint(c);

            c = scan();

        } while (!isWhiteSpace(c));

        return res.toString();

    }



    private boolean isWhiteSpace(int n) {

        if (n == ' ' || n == 'n' || n == 'r' || n == 't' || n == -1) return true;

        else return false;

    }



    public long scanLong() throws IOException {

        long integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public void scanLong(long A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanLong();

    }



    public void scanInt(int A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanInt();

    }



    public double scanDouble() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        int sgn = 1;

        if (c == '-') {

            sgn = -1;

            c = scan();

        }

        double res = 0;

        while (!isWhiteSpace(c) && c != '.') {

            if (c == 'e' || c == 'E') {

                return res * Math.pow(10, scanInt());

            }

            res *= 10;

            res += c - '0';

            c = scan();

        }

        if (c == '.') {

            c = scan();

            double m = 1;

            while (!isWhiteSpace(c)) {

                if (c == 'e' || c == 'E') {

                    return res * Math.pow(10, scanInt());

                }

                m /= 10;

                res += (c - '0') * m;

                c = scan();

            }

        }

        return res * sgn;

    }



}

Advantages :

Scans Input faster than BufferReader

Reduces Time Complexity

Flushes Buffer for every next input

Methods :

scanChar() - scan single character

scanInt() - scan Integer value

scanLong() - scan Long value

scanString() - scan String value

scanDouble() - scan Double value

scanInt(int array) - scans complete Array(Integer)

scanLong(long array) - scans complete Array(Long)

Usage :

Copy the Given Code below your java code.

Initialise Object for Given Class

ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :

import java.io.BufferedInputStream; import java.io.IOException; import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy

Example :

import java.io.BufferedInputStream;

import java.io.IOException;

import java.io.InputStream;

class Main{

    public static void main(String... as) throws IOException{

        ScanReader sc = new ScanReader(System.in);

        int a=sc.scanInt();

        System.out.println(a);

    }

}

class ScanReader....

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

add a comment |

sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

add a comment |

If you want to read both strings and ints, a solution is to use two Scanners:

Scanner stringScanner = new Scanner(System.in);

Scanner intScanner = new Scanner(System.in);



intScanner.nextInt();

String s = stringScanner.nextLine(); // unaffected by previous nextInt()

System.out.println(s);



intScanner.close();

stringScanner.close();

answered Jun 20 '17 at 18:52

André Valenti

34029

add a comment |

I guess I'm pretty late to the party..

As previously stated, calling input.nextLine() after getting your int value will solve your problem. The reason why your code didn't work was because there was nothing else to store from your input (where you inputted the int) into string1. I'll just shed a little more light to the entire topic.

Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:

int firstNumber = input.nextInt();

int secondNumber = input.nextInt();

If we input the value below (as a single line of input)

54 234

The value of our firstNumber and secondNumber variable become 54 and 234 respectively. The reason why this works this way is because a new line feed (i.e n) IS NOT automatically generated when the nextInt() method takes in the values. It simply takes the "next int" and moves on. This is the same for the rest of the nextFoo() methods except nextLine().

nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".

Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.

I hope this helps.. Merry coding!

edited Jan 7 '18 at 10:40

answered Jan 7 '18 at 10:33

Taslim

1,22741525

what is a "new line feed"?

– Abcd
Jan 13 at 16:22

@Abcd A new line feed basically means 'starting from a new line'.

– Taslim
Jan 13 at 19:35

add a comment |

public static void main(String args) {

        Scanner scan = new Scanner(System.in);

        int i = scan.nextInt();

        scan.nextLine();

        double d = scan.nextDouble();

        scan.nextLine();

        String s = scan.nextLine();



        System.out.println("String: " + s);

        System.out.println("Double: " + d);

        System.out.println("Int: " + i);

    }

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

– Neeraj Gahlawat
Jul 30 '17 at 3:50

add a comment |

-1

Use 2 scanner objects instead of one

Scanner input = new Scanner(System.in);

System.out.println("Enter numerical value");    

int option;

Scanner input2 = new Scanner(System.in);

option = input2.nextInt();

answered 8 hours ago

Harsh Shah

192

add a comment |

-2

Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.

int i = new Scanner(System.in).nextInt();

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

1

But then you have to close the Scanner to prevent memory leak. Wasting time?

– TheCoffeeCup
Oct 8 '15 at 1:22

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()};

– Tobias Johansson
Jun 1 '16 at 22:14

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one.

– Dawood ibn Kareem
Dec 29 '16 at 20:55

add a comment |

protected by Aniket Thakur Apr 19 '15 at 9:05

Thank you for your interest in this question.
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15 Answers
15

active

oldest

votes

15 Answers
15

active

oldest

votes

634

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
```
int option = input.nextInt();

input.nextLine();  // Consume newline left-over

String str1 = input.nextLine();
```

Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
```
int option = 0;

try {

    option = Integer.parseInt(input.nextLine());

} catch (NumberFormatException e) {

    e.printStackTrace();

}

String str1 = input.nextLine();
```

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above?

– blekione
Oct 27 '12 at 17:52

3

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right?

– Rohit Jain
Oct 27 '12 at 18:02

3

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions.

– Rohit Jain
Oct 27 '12 at 18:02

4

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign.

– Mordechai
Jan 11 '17 at 3:00

1

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too.

– MildlyMilquetoast
Jan 22 '17 at 1:13

|
show 4 more comments

634

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
```
int option = input.nextInt();

input.nextLine();  // Consume newline left-over

String str1 = input.nextLine();
```

Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
```
int option = 0;

try {

    option = Integer.parseInt(input.nextLine());

} catch (NumberFormatException e) {

    e.printStackTrace();

}

String str1 = input.nextLine();
```

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above?

– blekione
Oct 27 '12 at 17:52

3

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right?

– Rohit Jain
Oct 27 '12 at 18:02

3

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions.

– Rohit Jain
Oct 27 '12 at 18:02

4

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign.

– Mordechai
Jan 11 '17 at 3:00

1

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too.

– MildlyMilquetoast
Jan 22 '17 at 1:13

|
show 4 more comments

634

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
```
int option = input.nextInt();

input.nextLine();  // Consume newline left-over

String str1 = input.nextLine();
```

Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
```
int option = 0;

try {

    option = Integer.parseInt(input.nextLine());

} catch (NumberFormatException e) {

    e.printStackTrace();

}

String str1 = input.nextLine();
```

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
```
int option = input.nextInt();

input.nextLine();  // Consume newline left-over

String str1 = input.nextLine();
```

Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
```
int option = 0;

try {

    option = Integer.parseInt(input.nextLine());

} catch (NumberFormatException e) {

    e.printStackTrace();

}

String str1 = input.nextLine();
```

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

edited Oct 31 '18 at 1:31

TechnicallyTrue

233

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

answered Oct 27 '12 at 16:39

Rohit Jain

165k33310440

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above?

– blekione
Oct 27 '12 at 17:52

3

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right?

– Rohit Jain
Oct 27 '12 at 18:02

3

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions.

– Rohit Jain
Oct 27 '12 at 18:02

4

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign.

– Mordechai
Jan 11 '17 at 3:00

1

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too.

– MildlyMilquetoast
Jan 22 '17 at 1:13

|
show 4 more comments

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above?

– blekione
Oct 27 '12 at 17:52

3

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right?

– Rohit Jain
Oct 27 '12 at 18:02

3

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions.

– Rohit Jain
Oct 27 '12 at 18:02

4

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign.

– Mordechai
Jan 11 '17 at 3:00

1

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too.

– MildlyMilquetoast
Jan 22 '17 at 1:13

Couldn't I use only option = Integer.parseInt(input.nextLine());} instead of full try{...}? Is there any weakness of using only above?

– blekione
Oct 27 '12 at 17:52

@blekione. You have to use try-catch, because Integer.parseInt throws NumberFormatException when an invalid argument is passed to it. You will learn about exception later on. For E.G: - Integer.parseInt("abc"). You don't want "abc" to get converted to int right?

– Rohit Jain
Oct 27 '12 at 18:02

@blekione. So, in the above case, your code will halt at that point, and you won't be able to continue the execution. With Exception Handling, you can handle such kind of conditions.

– Rohit Jain
Oct 27 '12 at 18:02

To which extent is the latter better? AFAIK Scanner#nextInt() is way more lenient in finding correct ints, by allowing group commas and locale prefixes and suffixes. Integer#parseInt() allows digits and decimal point only plus an optional sign.

– Mordechai
Jan 11 '17 at 3:00

I personally prefer a Scanner#hasNextFoo check beforehand instead of a try-catch, but that works too.

– MildlyMilquetoast
Jan 22 '17 at 1:13

|
show 4 more comments

179

Try it like that:

System.out.print("Insert a number: ");

int number = input.nextInt();

input.nextLine(); // This line you have to add (It consumes the n character)

System.out.print("Text1: ");

String text1 = input.nextLine();

System.out.print("Text2: ");

String text2 = input.nextLine();

edited Dec 29 '17 at 6:57

Ebony Maw

316314

answered Aug 14 '11 at 12:24

Prine

8,59773355

umm seems a solution, making the skipped line non-used nextLine, but I still need an explanation of this behaviour

– Eng.Fouad
Aug 14 '11 at 12:25

2

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:11

add a comment |

179

Try it like that:

System.out.print("Insert a number: ");

int number = input.nextInt();

input.nextLine(); // This line you have to add (It consumes the n character)

System.out.print("Text1: ");

String text1 = input.nextLine();

System.out.print("Text2: ");

String text2 = input.nextLine();

edited Dec 29 '17 at 6:57

Ebony Maw

316314

answered Aug 14 '11 at 12:24

Prine

8,59773355

umm seems a solution, making the skipped line non-used nextLine, but I still need an explanation of this behaviour

– Eng.Fouad
Aug 14 '11 at 12:25

2

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:11

add a comment |

179

Try it like that:

System.out.print("Insert a number: ");

int number = input.nextInt();

input.nextLine(); // This line you have to add (It consumes the n character)

System.out.print("Text1: ");

String text1 = input.nextLine();

System.out.print("Text2: ");

String text2 = input.nextLine();

edited Dec 29 '17 at 6:57

Ebony Maw

316314

answered Aug 14 '11 at 12:24

Prine

8,59773355

Try it like that:

System.out.print("Insert a number: ");

int number = input.nextInt();

input.nextLine(); // This line you have to add (It consumes the n character)

System.out.print("Text1: ");

String text1 = input.nextLine();

System.out.print("Text2: ");

String text2 = input.nextLine();

edited Dec 29 '17 at 6:57

Ebony Maw

316314

answered Aug 14 '11 at 12:24

Prine

8,59773355

edited Dec 29 '17 at 6:57

Ebony Maw

316314

edited Dec 29 '17 at 6:57

Ebony Maw

316314

edited Dec 29 '17 at 6:57

Ebony Maw

316314

answered Aug 14 '11 at 12:24

Prine

8,59773355

answered Aug 14 '11 at 12:24

Prine

8,59773355

answered Aug 14 '11 at 12:24

Prine

8,59773355

umm seems a solution, making the skipped line non-used nextLine, but I still need an explanation of this behaviour

– Eng.Fouad
Aug 14 '11 at 12:25

2

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:11

add a comment |

umm seems a solution, making the skipped line non-used nextLine, but I still need an explanation of this behaviour

– Eng.Fouad
Aug 14 '11 at 12:25

2

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:11

umm seems a solution, making the skipped line non-used nextLine, but I still need an explanation of this behaviour

– Eng.Fouad
Aug 14 '11 at 12:25

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:11

add a comment |

Instead, use input.nextLine() immediately after input.nextInt()

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

Mmm, that bug was corrected in now a days?

– Victor
May 29 '14 at 13:28

7

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input.

– Bohemian♦
May 29 '14 at 13:38

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP.

– Victor
May 29 '14 at 14:55

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page.

– Bohemian♦
May 29 '14 at 15:18

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Instead, use input.nextLine() immediately after input.nextInt()

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

Mmm, that bug was corrected in now a days?

– Victor
May 29 '14 at 13:28

7

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input.

– Bohemian♦
May 29 '14 at 13:38

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP.

– Victor
May 29 '14 at 14:55

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page.

– Bohemian♦
May 29 '14 at 15:18

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Instead, use input.nextLine() immediately after input.nextInt()

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

Instead, use input.nextLine() immediately after input.nextInt()

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

answered Aug 14 '11 at 12:25

Bohemian♦

295k64415553

Mmm, that bug was corrected in now a days?

– Victor
May 29 '14 at 13:28

7

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input.

– Bohemian♦
May 29 '14 at 13:38

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP.

– Victor
May 29 '14 at 14:55

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page.

– Bohemian♦
May 29 '14 at 15:18

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Mmm, that bug was corrected in now a days?

– Victor
May 29 '14 at 13:28

7

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input.

– Bohemian♦
May 29 '14 at 13:38

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP.

– Victor
May 29 '14 at 14:55

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page.

– Bohemian♦
May 29 '14 at 15:18

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

Mmm, that bug was corrected in now a days?

– Victor
May 29 '14 at 13:28

@Victor it's not bug. it's working as specified. you could argue though that there should be an easy way to tell the api to also consume any whitespace following the target input.

– Bohemian♦
May 29 '14 at 13:38

I see. thanks @Bohemian, that exactly what i am arguing. I think that this should be changed, perhaps you can point me where to suggest this "issue" in the next JCP.

– Victor
May 29 '14 at 14:55

@victor You can request (and find out how to contribute code to) a feature via the Report a Bug or Request a Feature page.

– Bohemian♦
May 29 '14 at 15:18

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

edited May 23 '17 at 12:10

Community♦

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

I know what we do to remove the data in buffer, but this case, please help me with this: stackoverflow.com/questions/33585314/having-issues-with-scanner

– Khuong
Nov 7 '15 at 18:03

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: stackoverflow.com/a/42471816/7098259

– Patrick Parker
Feb 26 '17 at 17:52

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours.

– Denis Tulskiy
Feb 27 '17 at 5:57

@PatrickParker Why does is cause infinite loop ?

– Piyush
Aug 19 '17 at 8:41

|
show 1 more comment

edited May 23 '17 at 12:10

Community♦

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

I know what we do to remove the data in buffer, but this case, please help me with this: stackoverflow.com/questions/33585314/having-issues-with-scanner

– Khuong
Nov 7 '15 at 18:03

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: stackoverflow.com/a/42471816/7098259

– Patrick Parker
Feb 26 '17 at 17:52

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours.

– Denis Tulskiy
Feb 27 '17 at 5:57

@PatrickParker Why does is cause infinite loop ?

– Piyush
Aug 19 '17 at 8:41

|
show 1 more comment

edited May 23 '17 at 12:10

Community♦

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

edited May 23 '17 at 12:10

Community♦

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

edited May 23 '17 at 12:10

Community♦

edited May 23 '17 at 12:10

Community♦

edited May 23 '17 at 12:10

Community♦

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

answered Mar 23 '14 at 16:35

Denis Tulskiy

16.2k54261

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

I know what we do to remove the data in buffer, but this case, please help me with this: stackoverflow.com/questions/33585314/having-issues-with-scanner

– Khuong
Nov 7 '15 at 18:03

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: stackoverflow.com/a/42471816/7098259

– Patrick Parker
Feb 26 '17 at 17:52

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours.

– Denis Tulskiy
Feb 27 '17 at 5:57

@PatrickParker Why does is cause infinite loop ?

– Piyush
Aug 19 '17 at 8:41

|
show 1 more comment

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

I know what we do to remove the data in buffer, but this case, please help me with this: stackoverflow.com/questions/33585314/having-issues-with-scanner

– Khuong
Nov 7 '15 at 18:03

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: stackoverflow.com/a/42471816/7098259

– Patrick Parker
Feb 26 '17 at 17:52

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours.

– Denis Tulskiy
Feb 27 '17 at 5:57

@PatrickParker Why does is cause infinite loop ?

– Piyush
Aug 19 '17 at 8:41

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

I know what we do to remove the data in buffer, but this case, please help me with this: stackoverflow.com/questions/33585314/having-issues-with-scanner

– Khuong
Nov 7 '15 at 18:03

FYI, this will cause an infinite loop is the user types any tabs or spaces after the numeric input. See my answer here for a better skip: stackoverflow.com/a/42471816/7098259

– Patrick Parker
Feb 26 '17 at 17:52

@PatrickParker: it does indeed cause an infinite loop! Thanks, edited the answer to link to yours.

– Denis Tulskiy
Feb 27 '17 at 5:57

@PatrickParker Why does is cause infinite loop ?

– Piyush
Aug 19 '17 at 8:41

|
show 1 more comment

int number = Integer.parseInt(input.nextLine());

Doing this works just as well, and it saves you a line of code.

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

int number = Integer.parseInt(input.nextLine());

Doing this works just as well, and it saves you a line of code.

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

int number = Integer.parseInt(input.nextLine());

Doing this works just as well, and it saves you a line of code.

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

int number = Integer.parseInt(input.nextLine());

Doing this works just as well, and it saves you a line of code.

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

answered Feb 23 '13 at 22:01

Electric Coffee

5,63523996

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Things you need to know:

text which represents few lines also contains non-printable characters between lines (we call them line separators) like
- carriage return (CR - in String literals represented as "r")
- line feed (LF - in String literals represented as "n")

when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows rn) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42rn".

Problem

Solution

So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either

consume line separator left by nextInt from Scanners cache by
- calling nextLine,
- or by calling skip("\R") to let Scanner skip part matched by R which represents line separator (more info about R: https://stackoverflow.com/a/31060125)

don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.

int num1 = sc.nextInt();

int num2 = sc.nextInt();

String name = sc.next();

will be able to properly assign num1=42 num2=321 name=foobar.

edited Aug 15 '18 at 16:29

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

add a comment |

Things you need to know:

text which represents few lines also contains non-printable characters between lines (we call them line separators) like
- carriage return (CR - in String literals represented as "r")
- line feed (LF - in String literals represented as "n")

when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows rn) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42rn".

Problem

Solution

So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either

consume line separator left by nextInt from Scanners cache by
- calling nextLine,
- or by calling skip("\R") to let Scanner skip part matched by R which represents line separator (more info about R: https://stackoverflow.com/a/31060125)

don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.

int num1 = sc.nextInt();

int num2 = sc.nextInt();

String name = sc.next();

will be able to properly assign num1=42 num2=321 name=foobar.

edited Aug 15 '18 at 16:29

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

add a comment |

Things you need to know:

text which represents few lines also contains non-printable characters between lines (we call them line separators) like
- carriage return (CR - in String literals represented as "r")
- line feed (LF - in String literals represented as "n")

when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows rn) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42rn".

Problem

Solution

So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either

consume line separator left by nextInt from Scanners cache by
- calling nextLine,
- or by calling skip("\R") to let Scanner skip part matched by R which represents line separator (more info about R: https://stackoverflow.com/a/31060125)

don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.

int num1 = sc.nextInt();

int num2 = sc.nextInt();

String name = sc.next();

will be able to properly assign num1=42 num2=321 name=foobar.

edited Aug 15 '18 at 16:29

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

Things you need to know:

text which represents few lines also contains non-printable characters between lines (we call them line separators) like
- carriage return (CR - in String literals represented as "r")
- line feed (LF - in String literals represented as "n")

when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows rn) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42rn".

Problem

Solution

So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either

consume line separator left by nextInt from Scanners cache by
- calling nextLine,
- or by calling skip("\R") to let Scanner skip part matched by R which represents line separator (more info about R: https://stackoverflow.com/a/31060125)

don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.

int num1 = sc.nextInt();

int num2 = sc.nextInt();

String name = sc.next();

will be able to properly assign num1=42 num2=321 name=foobar.

edited Aug 15 '18 at 16:29

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

edited Aug 15 '18 at 16:29

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

answered Oct 9 '16 at 22:51

Pshemo

94.1k15130188

add a comment |

Instead of input.nextLine() use input.next(), that should solve the problem.

Modified code:

public static Scanner input = new Scanner(System.in);



public static void main(String args)

{

    System.out.print("Insert a number: ");

    int number = input.nextInt();

    System.out.print("Text1: ");

    String text1 = input.next();

    System.out.print("Text2: ");

    String text2 = input.next();

}

edited Jul 23 '14 at 10:31

arghtype

3,04192846

answered Jul 23 '14 at 10:20

Castaldi

374417

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Instead of input.nextLine() use input.next(), that should solve the problem.

Modified code:

public static Scanner input = new Scanner(System.in);



public static void main(String args)

{

    System.out.print("Insert a number: ");

    int number = input.nextInt();

    System.out.print("Text1: ");

    String text1 = input.next();

    System.out.print("Text2: ");

    String text2 = input.next();

}

edited Jul 23 '14 at 10:31

arghtype

3,04192846

answered Jul 23 '14 at 10:20

Castaldi

374417

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Instead of input.nextLine() use input.next(), that should solve the problem.

Modified code:

public static Scanner input = new Scanner(System.in);



public static void main(String args)

{

    System.out.print("Insert a number: ");

    int number = input.nextInt();

    System.out.print("Text1: ");

    String text1 = input.next();

    System.out.print("Text2: ");

    String text2 = input.next();

}

edited Jul 23 '14 at 10:31

arghtype

3,04192846

answered Jul 23 '14 at 10:20

Castaldi

374417

Instead of input.nextLine() use input.next(), that should solve the problem.

Modified code:

public static Scanner input = new Scanner(System.in);



public static void main(String args)

{

    System.out.print("Insert a number: ");

    int number = input.nextInt();

    System.out.print("Text1: ");

    String text1 = input.next();

    System.out.print("Text2: ");

    String text2 = input.next();

}

edited Jul 23 '14 at 10:31

arghtype

3,04192846

answered Jul 23 '14 at 10:20

Castaldi

374417

edited Jul 23 '14 at 10:31

arghtype

3,04192846

edited Jul 23 '14 at 10:31

arghtype

3,04192846

edited Jul 23 '14 at 10:31

arghtype

3,04192846

answered Jul 23 '14 at 10:20

Castaldi

374417

answered Jul 23 '14 at 10:20

Castaldi

374417

answered Jul 23 '14 at 10:20

Castaldi

374417

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

FYI: merged from stackoverflow.com/questions/7056749/…

– Shog9♦
Nov 13 '14 at 19:12

add a comment |

Scanner scanner =  new Scanner(System.in);

int option = scanner.nextInt();

scanner.nextLine(); //clearing the buffer

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

add a comment |

Scanner scanner =  new Scanner(System.in);

int option = scanner.nextInt();

scanner.nextLine(); //clearing the buffer

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

add a comment |

Scanner scanner =  new Scanner(System.in);

int option = scanner.nextInt();

scanner.nextLine(); //clearing the buffer

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

Scanner scanner =  new Scanner(System.in);

int option = scanner.nextInt();

scanner.nextLine(); //clearing the buffer

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

answered Nov 17 '16 at 0:51

Urvashi Gupta

24739

add a comment |

If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .

Code :

class ScanReader {

/**

* @author Nikunj Khokhar

*/

    private byte buf = new byte[4 * 1024];

    private int index;

    private BufferedInputStream in;

    private int total;



    public ScanReader(InputStream inputStream) {

        in = new BufferedInputStream(inputStream);

    }



    private int scan() throws IOException {

        if (index >= total) {

            index = 0;

            total = in.read(buf);

            if (total <= 0) return -1;

        }

        return buf[index++];

    }

    public char scanChar(){

        int c=scan();

        while (isWhiteSpace(c))c=scan();

        return (char)c;

    }





    public int scanInt() throws IOException {

        int integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public String scanString() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        StringBuilder res = new StringBuilder();

        do {

            res.appendCodePoint(c);

            c = scan();

        } while (!isWhiteSpace(c));

        return res.toString();

    }



    private boolean isWhiteSpace(int n) {

        if (n == ' ' || n == 'n' || n == 'r' || n == 't' || n == -1) return true;

        else return false;

    }



    public long scanLong() throws IOException {

        long integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public void scanLong(long A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanLong();

    }



    public void scanInt(int A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanInt();

    }



    public double scanDouble() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        int sgn = 1;

        if (c == '-') {

            sgn = -1;

            c = scan();

        }

        double res = 0;

        while (!isWhiteSpace(c) && c != '.') {

            if (c == 'e' || c == 'E') {

                return res * Math.pow(10, scanInt());

            }

            res *= 10;

            res += c - '0';

            c = scan();

        }

        if (c == '.') {

            c = scan();

            double m = 1;

            while (!isWhiteSpace(c)) {

                if (c == 'e' || c == 'E') {

                    return res * Math.pow(10, scanInt());

                }

                m /= 10;

                res += (c - '0') * m;

                c = scan();

            }

        }

        return res * sgn;

    }



}

Advantages :

Scans Input faster than BufferReader

Reduces Time Complexity

Flushes Buffer for every next input

Methods :

scanChar() - scan single character

scanInt() - scan Integer value

scanLong() - scan Long value

scanString() - scan String value

scanDouble() - scan Double value

scanInt(int array) - scans complete Array(Integer)

scanLong(long array) - scans complete Array(Long)

Usage :

Copy the Given Code below your java code.

Initialise Object for Given Class

ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :

import java.io.BufferedInputStream; import java.io.IOException; import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy

Example :

import java.io.BufferedInputStream;

import java.io.IOException;

import java.io.InputStream;

class Main{

    public static void main(String... as) throws IOException{

        ScanReader sc = new ScanReader(System.in);

        int a=sc.scanInt();

        System.out.println(a);

    }

}

class ScanReader....

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

add a comment |

If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .

Code :

class ScanReader {

/**

* @author Nikunj Khokhar

*/

    private byte buf = new byte[4 * 1024];

    private int index;

    private BufferedInputStream in;

    private int total;



    public ScanReader(InputStream inputStream) {

        in = new BufferedInputStream(inputStream);

    }



    private int scan() throws IOException {

        if (index >= total) {

            index = 0;

            total = in.read(buf);

            if (total <= 0) return -1;

        }

        return buf[index++];

    }

    public char scanChar(){

        int c=scan();

        while (isWhiteSpace(c))c=scan();

        return (char)c;

    }





    public int scanInt() throws IOException {

        int integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public String scanString() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        StringBuilder res = new StringBuilder();

        do {

            res.appendCodePoint(c);

            c = scan();

        } while (!isWhiteSpace(c));

        return res.toString();

    }



    private boolean isWhiteSpace(int n) {

        if (n == ' ' || n == 'n' || n == 'r' || n == 't' || n == -1) return true;

        else return false;

    }



    public long scanLong() throws IOException {

        long integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public void scanLong(long A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanLong();

    }



    public void scanInt(int A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanInt();

    }



    public double scanDouble() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        int sgn = 1;

        if (c == '-') {

            sgn = -1;

            c = scan();

        }

        double res = 0;

        while (!isWhiteSpace(c) && c != '.') {

            if (c == 'e' || c == 'E') {

                return res * Math.pow(10, scanInt());

            }

            res *= 10;

            res += c - '0';

            c = scan();

        }

        if (c == '.') {

            c = scan();

            double m = 1;

            while (!isWhiteSpace(c)) {

                if (c == 'e' || c == 'E') {

                    return res * Math.pow(10, scanInt());

                }

                m /= 10;

                res += (c - '0') * m;

                c = scan();

            }

        }

        return res * sgn;

    }



}

Advantages :

Scans Input faster than BufferReader

Reduces Time Complexity

Flushes Buffer for every next input

Methods :

scanChar() - scan single character

scanInt() - scan Integer value

scanLong() - scan Long value

scanString() - scan String value

scanDouble() - scan Double value

scanInt(int array) - scans complete Array(Integer)

scanLong(long array) - scans complete Array(Long)

Usage :

Copy the Given Code below your java code.

Initialise Object for Given Class

ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :

import java.io.BufferedInputStream; import java.io.IOException; import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy

Example :

import java.io.BufferedInputStream;

import java.io.IOException;

import java.io.InputStream;

class Main{

    public static void main(String... as) throws IOException{

        ScanReader sc = new ScanReader(System.in);

        int a=sc.scanInt();

        System.out.println(a);

    }

}

class ScanReader....

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

add a comment |

If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .

Code :

class ScanReader {

/**

* @author Nikunj Khokhar

*/

    private byte buf = new byte[4 * 1024];

    private int index;

    private BufferedInputStream in;

    private int total;



    public ScanReader(InputStream inputStream) {

        in = new BufferedInputStream(inputStream);

    }



    private int scan() throws IOException {

        if (index >= total) {

            index = 0;

            total = in.read(buf);

            if (total <= 0) return -1;

        }

        return buf[index++];

    }

    public char scanChar(){

        int c=scan();

        while (isWhiteSpace(c))c=scan();

        return (char)c;

    }





    public int scanInt() throws IOException {

        int integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public String scanString() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        StringBuilder res = new StringBuilder();

        do {

            res.appendCodePoint(c);

            c = scan();

        } while (!isWhiteSpace(c));

        return res.toString();

    }



    private boolean isWhiteSpace(int n) {

        if (n == ' ' || n == 'n' || n == 'r' || n == 't' || n == -1) return true;

        else return false;

    }



    public long scanLong() throws IOException {

        long integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public void scanLong(long A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanLong();

    }



    public void scanInt(int A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanInt();

    }



    public double scanDouble() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        int sgn = 1;

        if (c == '-') {

            sgn = -1;

            c = scan();

        }

        double res = 0;

        while (!isWhiteSpace(c) && c != '.') {

            if (c == 'e' || c == 'E') {

                return res * Math.pow(10, scanInt());

            }

            res *= 10;

            res += c - '0';

            c = scan();

        }

        if (c == '.') {

            c = scan();

            double m = 1;

            while (!isWhiteSpace(c)) {

                if (c == 'e' || c == 'E') {

                    return res * Math.pow(10, scanInt());

                }

                m /= 10;

                res += (c - '0') * m;

                c = scan();

            }

        }

        return res * sgn;

    }



}

Advantages :

Scans Input faster than BufferReader

Reduces Time Complexity

Flushes Buffer for every next input

Methods :

scanChar() - scan single character

scanInt() - scan Integer value

scanLong() - scan Long value

scanString() - scan String value

scanDouble() - scan Double value

scanInt(int array) - scans complete Array(Integer)

scanLong(long array) - scans complete Array(Long)

Usage :

Copy the Given Code below your java code.

Initialise Object for Given Class

ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :

import java.io.BufferedInputStream; import java.io.IOException; import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy

Example :

import java.io.BufferedInputStream;

import java.io.IOException;

import java.io.InputStream;

class Main{

    public static void main(String... as) throws IOException{

        ScanReader sc = new ScanReader(System.in);

        int a=sc.scanInt();

        System.out.println(a);

    }

}

class ScanReader....

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .

Code :

class ScanReader {

/**

* @author Nikunj Khokhar

*/

    private byte buf = new byte[4 * 1024];

    private int index;

    private BufferedInputStream in;

    private int total;



    public ScanReader(InputStream inputStream) {

        in = new BufferedInputStream(inputStream);

    }



    private int scan() throws IOException {

        if (index >= total) {

            index = 0;

            total = in.read(buf);

            if (total <= 0) return -1;

        }

        return buf[index++];

    }

    public char scanChar(){

        int c=scan();

        while (isWhiteSpace(c))c=scan();

        return (char)c;

    }





    public int scanInt() throws IOException {

        int integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public String scanString() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        StringBuilder res = new StringBuilder();

        do {

            res.appendCodePoint(c);

            c = scan();

        } while (!isWhiteSpace(c));

        return res.toString();

    }



    private boolean isWhiteSpace(int n) {

        if (n == ' ' || n == 'n' || n == 'r' || n == 't' || n == -1) return true;

        else return false;

    }



    public long scanLong() throws IOException {

        long integer = 0;

        int n = scan();

        while (isWhiteSpace(n)) n = scan();

        int neg = 1;

        if (n == '-') {

            neg = -1;

            n = scan();

        }

        while (!isWhiteSpace(n)) {

            if (n >= '0' && n <= '9') {

                integer *= 10;

                integer += n - '0';

                n = scan();

            }

        }

        return neg * integer;

    }



    public void scanLong(long A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanLong();

    }



    public void scanInt(int A) throws IOException {

        for (int i = 0; i < A.length; i++) A[i] = scanInt();

    }



    public double scanDouble() throws IOException {

        int c = scan();

        while (isWhiteSpace(c)) c = scan();

        int sgn = 1;

        if (c == '-') {

            sgn = -1;

            c = scan();

        }

        double res = 0;

        while (!isWhiteSpace(c) && c != '.') {

            if (c == 'e' || c == 'E') {

                return res * Math.pow(10, scanInt());

            }

            res *= 10;

            res += c - '0';

            c = scan();

        }

        if (c == '.') {

            c = scan();

            double m = 1;

            while (!isWhiteSpace(c)) {

                if (c == 'e' || c == 'E') {

                    return res * Math.pow(10, scanInt());

                }

                m /= 10;

                res += (c - '0') * m;

                c = scan();

            }

        }

        return res * sgn;

    }



}

Advantages :

Scans Input faster than BufferReader

Reduces Time Complexity

Flushes Buffer for every next input

Methods :

scanChar() - scan single character

scanInt() - scan Integer value

scanLong() - scan Long value

scanString() - scan String value

scanDouble() - scan Double value

scanInt(int array) - scans complete Array(Integer)

scanLong(long array) - scans complete Array(Long)

Usage :

Copy the Given Code below your java code.

Initialise Object for Given Class

ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :

import java.io.BufferedInputStream; import java.io.IOException; import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy

Example :

import java.io.BufferedInputStream;

import java.io.IOException;

import java.io.InputStream;

class Main{

    public static void main(String... as) throws IOException{

        ScanReader sc = new ScanReader(System.in);

        int a=sc.scanInt();

        System.out.println(a);

    }

}

class ScanReader....

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

answered Jun 16 '17 at 8:12

NIKUNJ KHOKHAR

311313

add a comment |

sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

add a comment |

sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

add a comment |

sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

edited Sep 14 '17 at 10:45

Aurasphere

2,519102850

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

answered Sep 14 '17 at 10:16

shankar Dayal upadhyay

15028

add a comment |

If you want to read both strings and ints, a solution is to use two Scanners:

Scanner stringScanner = new Scanner(System.in);

Scanner intScanner = new Scanner(System.in);



intScanner.nextInt();

String s = stringScanner.nextLine(); // unaffected by previous nextInt()

System.out.println(s);



intScanner.close();

stringScanner.close();

answered Jun 20 '17 at 18:52

André Valenti

34029

add a comment |

If you want to read both strings and ints, a solution is to use two Scanners:

Scanner stringScanner = new Scanner(System.in);

Scanner intScanner = new Scanner(System.in);



intScanner.nextInt();

String s = stringScanner.nextLine(); // unaffected by previous nextInt()

System.out.println(s);



intScanner.close();

stringScanner.close();

answered Jun 20 '17 at 18:52

André Valenti

34029

add a comment |

If you want to read both strings and ints, a solution is to use two Scanners:

Scanner stringScanner = new Scanner(System.in);

Scanner intScanner = new Scanner(System.in);



intScanner.nextInt();

String s = stringScanner.nextLine(); // unaffected by previous nextInt()

System.out.println(s);



intScanner.close();

stringScanner.close();

answered Jun 20 '17 at 18:52

André Valenti

34029

If you want to read both strings and ints, a solution is to use two Scanners:

Scanner stringScanner = new Scanner(System.in);

Scanner intScanner = new Scanner(System.in);



intScanner.nextInt();

String s = stringScanner.nextLine(); // unaffected by previous nextInt()

System.out.println(s);



intScanner.close();

stringScanner.close();

answered Jun 20 '17 at 18:52

André Valenti

34029

answered Jun 20 '17 at 18:52

André Valenti

34029

answered Jun 20 '17 at 18:52

André Valenti

34029

answered Jun 20 '17 at 18:52

André Valenti

34029

add a comment |

I guess I'm pretty late to the party..

Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:

int firstNumber = input.nextInt();

int secondNumber = input.nextInt();

If we input the value below (as a single line of input)

54 234

nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".

Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.

I hope this helps.. Merry coding!

edited Jan 7 '18 at 10:40

answered Jan 7 '18 at 10:33

Taslim

1,22741525

what is a "new line feed"?

– Abcd
Jan 13 at 16:22

@Abcd A new line feed basically means 'starting from a new line'.

– Taslim
Jan 13 at 19:35

add a comment |

I guess I'm pretty late to the party..

Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:

int firstNumber = input.nextInt();

int secondNumber = input.nextInt();

If we input the value below (as a single line of input)

54 234

nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".

Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.

I hope this helps.. Merry coding!

edited Jan 7 '18 at 10:40

answered Jan 7 '18 at 10:33

Taslim

1,22741525

what is a "new line feed"?

– Abcd
Jan 13 at 16:22

@Abcd A new line feed basically means 'starting from a new line'.

– Taslim
Jan 13 at 19:35

add a comment |

I guess I'm pretty late to the party..

Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:

int firstNumber = input.nextInt();

int secondNumber = input.nextInt();

If we input the value below (as a single line of input)

54 234

nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".

Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.

I hope this helps.. Merry coding!

edited Jan 7 '18 at 10:40

answered Jan 7 '18 at 10:33

Taslim

1,22741525

I guess I'm pretty late to the party..

Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:

int firstNumber = input.nextInt();

int secondNumber = input.nextInt();

If we input the value below (as a single line of input)

54 234

nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".

Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.

I hope this helps.. Merry coding!

edited Jan 7 '18 at 10:40

answered Jan 7 '18 at 10:33

Taslim

1,22741525

edited Jan 7 '18 at 10:40

answered Jan 7 '18 at 10:33

Taslim

1,22741525

answered Jan 7 '18 at 10:33

Taslim

1,22741525

answered Jan 7 '18 at 10:33

Taslim

1,22741525

what is a "new line feed"?

– Abcd
Jan 13 at 16:22

@Abcd A new line feed basically means 'starting from a new line'.

– Taslim
Jan 13 at 19:35

add a comment |

what is a "new line feed"?

– Abcd
Jan 13 at 16:22

@Abcd A new line feed basically means 'starting from a new line'.

– Taslim
Jan 13 at 19:35

what is a "new line feed"?

– Abcd
Jan 13 at 16:22

@Abcd A new line feed basically means 'starting from a new line'.

– Taslim
Jan 13 at 19:35

add a comment |

public static void main(String args) {

        Scanner scan = new Scanner(System.in);

        int i = scan.nextInt();

        scan.nextLine();

        double d = scan.nextDouble();

        scan.nextLine();

        String s = scan.nextLine();



        System.out.println("String: " + s);

        System.out.println("Double: " + d);

        System.out.println("Int: " + i);

    }

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

– Neeraj Gahlawat
Jul 30 '17 at 3:50

add a comment |

public static void main(String args) {

        Scanner scan = new Scanner(System.in);

        int i = scan.nextInt();

        scan.nextLine();

        double d = scan.nextDouble();

        scan.nextLine();

        String s = scan.nextLine();



        System.out.println("String: " + s);

        System.out.println("Double: " + d);

        System.out.println("Int: " + i);

    }

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

– Neeraj Gahlawat
Jul 30 '17 at 3:50

add a comment |

public static void main(String args) {

        Scanner scan = new Scanner(System.in);

        int i = scan.nextInt();

        scan.nextLine();

        double d = scan.nextDouble();

        scan.nextLine();

        String s = scan.nextLine();



        System.out.println("String: " + s);

        System.out.println("Double: " + d);

        System.out.println("Int: " + i);

    }

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

public static void main(String args) {

        Scanner scan = new Scanner(System.in);

        int i = scan.nextInt();

        scan.nextLine();

        double d = scan.nextDouble();

        scan.nextLine();

        String s = scan.nextLine();



        System.out.println("String: " + s);

        System.out.println("Double: " + d);

        System.out.println("Int: " + i);

    }

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

answered Jul 30 '17 at 3:48

Neeraj Gahlawat

45342

If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

– Neeraj Gahlawat
Jul 30 '17 at 3:50

add a comment |

If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

– Neeraj Gahlawat
Jul 30 '17 at 3:50

If you use the nextLine() method immediately following the nextInt() method, recall that nextInt() reads integer tokens; because of this, the last newline character for that line of integer input is still queued in the input buffer and the next nextLine() will be reading the remainder of the integer line (which is empty).

– Neeraj Gahlawat
Jul 30 '17 at 3:50

add a comment |

-1

Use 2 scanner objects instead of one

Scanner input = new Scanner(System.in);

System.out.println("Enter numerical value");    

int option;

Scanner input2 = new Scanner(System.in);

option = input2.nextInt();

answered 8 hours ago

Harsh Shah

192

add a comment |

-1

Use 2 scanner objects instead of one

Scanner input = new Scanner(System.in);

System.out.println("Enter numerical value");    

int option;

Scanner input2 = new Scanner(System.in);

option = input2.nextInt();

answered 8 hours ago

Harsh Shah

192

add a comment |

-1

Use 2 scanner objects instead of one

Scanner input = new Scanner(System.in);

System.out.println("Enter numerical value");    

int option;

Scanner input2 = new Scanner(System.in);

option = input2.nextInt();

answered 8 hours ago

Harsh Shah

192

Use 2 scanner objects instead of one

Scanner input = new Scanner(System.in);

System.out.println("Enter numerical value");    

int option;

Scanner input2 = new Scanner(System.in);

option = input2.nextInt();

answered 8 hours ago

Harsh Shah

192

answered 8 hours ago

Harsh Shah

192

answered 8 hours ago

Harsh Shah

192

answered 8 hours ago

Harsh Shah

192

add a comment |

-2

Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.

int i = new Scanner(System.in).nextInt();

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

1

But then you have to close the Scanner to prevent memory leak. Wasting time?

– TheCoffeeCup
Oct 8 '15 at 1:22

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()};

– Tobias Johansson
Jun 1 '16 at 22:14

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one.

– Dawood ibn Kareem
Dec 29 '16 at 20:55

add a comment |

-2

Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.

int i = new Scanner(System.in).nextInt();

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

1

But then you have to close the Scanner to prevent memory leak. Wasting time?

– TheCoffeeCup
Oct 8 '15 at 1:22

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()};

– Tobias Johansson
Jun 1 '16 at 22:14

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one.

– Dawood ibn Kareem
Dec 29 '16 at 20:55

add a comment |

-2

Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.

int i = new Scanner(System.in).nextInt();

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.

int i = new Scanner(System.in).nextInt();

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

answered Mar 13 '15 at 11:06

Tobias Johansson

170111

1

But then you have to close the Scanner to prevent memory leak. Wasting time?

– TheCoffeeCup
Oct 8 '15 at 1:22

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()};

– Tobias Johansson
Jun 1 '16 at 22:14

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one.

– Dawood ibn Kareem
Dec 29 '16 at 20:55

add a comment |

1

But then you have to close the Scanner to prevent memory leak. Wasting time?

– TheCoffeeCup
Oct 8 '15 at 1:22

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()};

– Tobias Johansson
Jun 1 '16 at 22:14

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one.

– Dawood ibn Kareem
Dec 29 '16 at 20:55

But then you have to close the Scanner to prevent memory leak. Wasting time?

– TheCoffeeCup
Oct 8 '15 at 1:22

Doesn't the GC take care of that if the Scanner is wrapped in a method? Like: String getInput() {return new Scanner(System.in).nextLine()};

– Tobias Johansson
Jun 1 '16 at 22:14

This is absolutely incorrect. nextInt() won't consume the newline, regardless of whether it's in a "new" Scanner or an already used one.

– Dawood ibn Kareem
Dec 29 '16 at 20:55

add a comment |

protected by Aniket Thakur Apr 19 '15 at 9:05

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