How read two lines numbers from input to 2 arrays, C language? Without knowing the amount of number? in C

For example, input is:

12 23 32 41 45

22 11 43

lines end with 'n',

I want save nums to a and b;

a = {12, 23, 32, 41, 45}

b = {22, 11, 43}

The point is I DON'T KNOW how many number of each line.

If I know line1 n numbers and line2 m numbers, I will use "for loop", and no question.

Just like,

for(int i = 0; i < n; i++) scanf("%d", a+i);

for(int i = 0; i < m; i++) scanf("%d", b+i);

But, I DO NOT know n and m, How to do, guys?

edited Nov 19 '18 at 19:54

ReAl

6061216

asked Nov 19 '18 at 16:58

user7331489

Can you accept a number max for array length ?
– Ôrel
Nov 19 '18 at 17:03

1

If you care about line endings, then you either need to read one character at a time with fgetc, or read a line at a time with fgets.
– user3386109
Nov 19 '18 at 17:06

Read characters until a new line is reached, if one isn't reached, resize your buffer? You'll want to use dynamic memory allocation probably.
– jacob
Nov 19 '18 at 17:07

2

A good function for reading a line at a time is getline, which was standardized in POSIX.1-2008. In comparision to fgets, it can read a line of arbitrary length into a dynamically allocated buffer. If you don't have the getline function, there are various standalone implementations of it, such as the one at github.com/ivanrad/getline .
– Ian Abbott
Nov 19 '18 at 17:19

You could use a while loop and some arbitrary input as break condition & dynamically alloc your input array(i think).
– ats
Nov 19 '18 at 17:25

|
show 1 more comment

For example, input is:

12 23 32 41 45

22 11 43

lines end with 'n',

I want save nums to a and b;

a = {12, 23, 32, 41, 45}

b = {22, 11, 43}

The point is I DON'T KNOW how many number of each line.

If I know line1 n numbers and line2 m numbers, I will use "for loop", and no question.

Just like,

for(int i = 0; i < n; i++) scanf("%d", a+i);

for(int i = 0; i < m; i++) scanf("%d", b+i);

But, I DO NOT know n and m, How to do, guys?

edited Nov 19 '18 at 19:54

ReAl

6061216

asked Nov 19 '18 at 16:58

user7331489

Can you accept a number max for array length ?
– Ôrel
Nov 19 '18 at 17:03

1

If you care about line endings, then you either need to read one character at a time with fgetc, or read a line at a time with fgets.
– user3386109
Nov 19 '18 at 17:06

Read characters until a new line is reached, if one isn't reached, resize your buffer? You'll want to use dynamic memory allocation probably.
– jacob
Nov 19 '18 at 17:07

2

A good function for reading a line at a time is getline, which was standardized in POSIX.1-2008. In comparision to fgets, it can read a line of arbitrary length into a dynamically allocated buffer. If you don't have the getline function, there are various standalone implementations of it, such as the one at github.com/ivanrad/getline .
– Ian Abbott
Nov 19 '18 at 17:19

You could use a while loop and some arbitrary input as break condition & dynamically alloc your input array(i think).
– ats
Nov 19 '18 at 17:25

|
show 1 more comment

For example, input is:

12 23 32 41 45

22 11 43

lines end with 'n',

I want save nums to a and b;

a = {12, 23, 32, 41, 45}

b = {22, 11, 43}

The point is I DON'T KNOW how many number of each line.

If I know line1 n numbers and line2 m numbers, I will use "for loop", and no question.

Just like,

for(int i = 0; i < n; i++) scanf("%d", a+i);

for(int i = 0; i < m; i++) scanf("%d", b+i);

But, I DO NOT know n and m, How to do, guys?

edited Nov 19 '18 at 19:54

ReAl

6061216

asked Nov 19 '18 at 16:58

user7331489

For example, input is:

12 23 32 41 45

22 11 43

lines end with 'n',

I want save nums to a and b;

a = {12, 23, 32, 41, 45}

b = {22, 11, 43}

The point is I DON'T KNOW how many number of each line.

If I know line1 n numbers and line2 m numbers, I will use "for loop", and no question.

Just like,

for(int i = 0; i < n; i++) scanf("%d", a+i);

for(int i = 0; i < m; i++) scanf("%d", b+i);

But, I DO NOT know n and m, How to do, guys?

c scanf

edited Nov 19 '18 at 19:54

ReAl

6061216

asked Nov 19 '18 at 16:58

user7331489

edited Nov 19 '18 at 19:54

ReAl

6061216

asked Nov 19 '18 at 16:58

user7331489

edited Nov 19 '18 at 19:54

ReAl

6061216

edited Nov 19 '18 at 19:54

ReAl

6061216

edited Nov 19 '18 at 19:54

ReAl

6061216

asked Nov 19 '18 at 16:58

user7331489

asked Nov 19 '18 at 16:58

user7331489

asked Nov 19 '18 at 16:58

user7331489

Can you accept a number max for array length ?
– Ôrel
Nov 19 '18 at 17:03

1

If you care about line endings, then you either need to read one character at a time with fgetc, or read a line at a time with fgets.
– user3386109
Nov 19 '18 at 17:06

Read characters until a new line is reached, if one isn't reached, resize your buffer? You'll want to use dynamic memory allocation probably.
– jacob
Nov 19 '18 at 17:07

2

A good function for reading a line at a time is getline, which was standardized in POSIX.1-2008. In comparision to fgets, it can read a line of arbitrary length into a dynamically allocated buffer. If you don't have the getline function, there are various standalone implementations of it, such as the one at github.com/ivanrad/getline .
– Ian Abbott
Nov 19 '18 at 17:19

You could use a while loop and some arbitrary input as break condition & dynamically alloc your input array(i think).
– ats
Nov 19 '18 at 17:25

|
show 1 more comment

Can you accept a number max for array length ?
– Ôrel
Nov 19 '18 at 17:03

1

If you care about line endings, then you either need to read one character at a time with fgetc, or read a line at a time with fgets.
– user3386109
Nov 19 '18 at 17:06

Read characters until a new line is reached, if one isn't reached, resize your buffer? You'll want to use dynamic memory allocation probably.
– jacob
Nov 19 '18 at 17:07

2

A good function for reading a line at a time is getline, which was standardized in POSIX.1-2008. In comparision to fgets, it can read a line of arbitrary length into a dynamically allocated buffer. If you don't have the getline function, there are various standalone implementations of it, such as the one at github.com/ivanrad/getline .
– Ian Abbott
Nov 19 '18 at 17:19

You could use a while loop and some arbitrary input as break condition & dynamically alloc your input array(i think).
– ats
Nov 19 '18 at 17:25

Can you accept a number max for array length ?
– Ôrel
Nov 19 '18 at 17:03

If you care about line endings, then you either need to read one character at a time with fgetc, or read a line at a time with fgets.
– user3386109
Nov 19 '18 at 17:06

Read characters until a new line is reached, if one isn't reached, resize your buffer? You'll want to use dynamic memory allocation probably.
– jacob
Nov 19 '18 at 17:07

A good function for reading a line at a time is getline, which was standardized in POSIX.1-2008. In comparision to fgets, it can read a line of arbitrary length into a dynamically allocated buffer. If you don't have the getline function, there are various standalone implementations of it, such as the one at github.com/ivanrad/getline .
– Ian Abbott
Nov 19 '18 at 17:19

You could use a while loop and some arbitrary input as break condition & dynamically alloc your input array(i think).
– ats
Nov 19 '18 at 17:25

|
show 1 more comment

3 Answers
3

active

oldest

votes

If you would like to continue using a scanf approach, I would recommend the negated scanset %[^] format specifier.

scanf("%[^n]", pointerToCharArray)

This should read in any number of characters up to but not including the character specified (which, in our case, is a newline). If you'd like to discard the newline, read it in as follows:

scanf("%[^n]n", pointerToCharArray)

A link to the reference page can be found below. The negated scanset specifier is included in the list:

http://www.cplusplus.com/reference/cstdio/scanf/

From this point, it is a simple matter to use strtok() to tokenize the output of the scanf into number arrays. If you are not familiar with strtok, another reference link is provided below:

http://www.cplusplus.com/reference/cstring/strtok/

I hope this helps!

answered Nov 19 '18 at 17:51

armitus

845

scanf("%[^n]n", pointerToCharArray) not only consumes a following newline, it waits for additional input consuming all other white-spaces. scanf() then does not return until following non-whitespace detected or EOF.
– chux
Nov 19 '18 at 18:57

As OP is concerned about "I DON'T KNOW how many number ", using "%[^n]" without a width limit is fragile code.
– chux
Nov 19 '18 at 18:58

Thank you very much!
– user7331489
Nov 22 '18 at 8:43

add a comment |

Let us use the non standard getline, a common library extension. @Abbott to read the entire line into allocated memory.

If an array is needed and its size determined at run-time, use a variable length array available in C99 and optionally in C11.

Make a function to count and maybe save the numbers. Use strto...() functions for robust parsing.

size_t count_longs(char *line, long *dest) {

  size_t count = 0;

  char *s = line;

  for (;;) {

    char *endptr;

    long num = strtol(s, &endptr, 10);

    if (s == endptr) {

      break;  /// no conversion

    }

    s = endptr;

    if (dest) {

      dest[count] = num;

    }

    count++;

  }

  return count;

}

Sample code snippet

  char *line = NULL;

  size_t len = 0;

  ssize_t nread = getline(&line, &len, stdin);  // read the whole line

  if (nread != -1) {

    // successful read



    long a[count_longs(line, NULL)];  // determine array size and use a 

    count_longs(line, a);             // 2nd pass: assign array elements. 



    nread = getline(&line, &len, stdin);

    if (nread != -1) {

      // successful read



      long b[count_longs(line, NULL)];

      count_longs(line, b);



      // Do something with a and b

    }

  }

  free(line);

  len = 0;

edited Nov 19 '18 at 18:19

answered Nov 19 '18 at 18:05

chux

80.8k870148

add a comment |

A solution that could potentially fit what OP asked for would be the code below.
The solution reads an unspecified number of integer values & stores them in arr. to break the loop/end input simply input a non-integer value. This could be further specified to search for a specific input etc.

int c = 0;

int *arr = NULL;

int *extArr = NULL;

int i = 0;

for (i = 0; scanf("%d", &c) == 1; i++)

{

    extArr = (int*)realloc(arr, (i+1) * sizeof(int)); 

    arr = extArr;

    arr[i] = c; 



}

free(arr);

I would, however personally go with a mix between the solution i provided & armitus answer(if i cant use getline or similar) since reallocating memory for every single int value seems redundant to me.

Edit Since there has been some question about the code i provided and how to best use it in order to fulfill the OP's question. The codesnippet was supposed to showcase how one might create his own scanf-style function for unspecified number of integer inputs reading one full array at a time(in that case the free() has to be omitted till one is done with the array(obviously).

edited Nov 19 '18 at 19:08

answered Nov 19 '18 at 18:37

ats

694

I do see a functional problem. scanf("%d", &c) will read through an end-of-line like any other white-space. I'd expect OP's two lines of integers to end up in one arr here.
– chux
Nov 19 '18 at 19:02

The idea was to use my codesnippet as a standalone function & then just call it n times for the number of array's required. i admittedly omitted those details from my answer.
– ats
Nov 19 '18 at 19:06

Trouble is we do not know n prior to reading the line.
– chux
Nov 19 '18 at 19:08

Maybe i should've phrased that better. I did not mean n as in array length, but n as in number of array's read. Since the codesnippet can easily be packed inside a function and used as a sort of scanf-overload for array's
– ats
Nov 19 '18 at 19:10

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

If you would like to continue using a scanf approach, I would recommend the negated scanset %[^] format specifier.

scanf("%[^n]", pointerToCharArray)

This should read in any number of characters up to but not including the character specified (which, in our case, is a newline). If you'd like to discard the newline, read it in as follows:

scanf("%[^n]n", pointerToCharArray)

A link to the reference page can be found below. The negated scanset specifier is included in the list:

http://www.cplusplus.com/reference/cstdio/scanf/

From this point, it is a simple matter to use strtok() to tokenize the output of the scanf into number arrays. If you are not familiar with strtok, another reference link is provided below:

http://www.cplusplus.com/reference/cstring/strtok/

I hope this helps!

answered Nov 19 '18 at 17:51

armitus

845

scanf("%[^n]n", pointerToCharArray) not only consumes a following newline, it waits for additional input consuming all other white-spaces. scanf() then does not return until following non-whitespace detected or EOF.
– chux
Nov 19 '18 at 18:57

As OP is concerned about "I DON'T KNOW how many number ", using "%[^n]" without a width limit is fragile code.
– chux
Nov 19 '18 at 18:58

Thank you very much!
– user7331489
Nov 22 '18 at 8:43

add a comment |

If you would like to continue using a scanf approach, I would recommend the negated scanset %[^] format specifier.

scanf("%[^n]", pointerToCharArray)

This should read in any number of characters up to but not including the character specified (which, in our case, is a newline). If you'd like to discard the newline, read it in as follows:

scanf("%[^n]n", pointerToCharArray)

A link to the reference page can be found below. The negated scanset specifier is included in the list:

http://www.cplusplus.com/reference/cstdio/scanf/

From this point, it is a simple matter to use strtok() to tokenize the output of the scanf into number arrays. If you are not familiar with strtok, another reference link is provided below:

http://www.cplusplus.com/reference/cstring/strtok/

I hope this helps!

answered Nov 19 '18 at 17:51

armitus

845

scanf("%[^n]n", pointerToCharArray) not only consumes a following newline, it waits for additional input consuming all other white-spaces. scanf() then does not return until following non-whitespace detected or EOF.
– chux
Nov 19 '18 at 18:57

As OP is concerned about "I DON'T KNOW how many number ", using "%[^n]" without a width limit is fragile code.
– chux
Nov 19 '18 at 18:58

Thank you very much!
– user7331489
Nov 22 '18 at 8:43

add a comment |

If you would like to continue using a scanf approach, I would recommend the negated scanset %[^] format specifier.

scanf("%[^n]", pointerToCharArray)

This should read in any number of characters up to but not including the character specified (which, in our case, is a newline). If you'd like to discard the newline, read it in as follows:

scanf("%[^n]n", pointerToCharArray)

A link to the reference page can be found below. The negated scanset specifier is included in the list:

http://www.cplusplus.com/reference/cstdio/scanf/

From this point, it is a simple matter to use strtok() to tokenize the output of the scanf into number arrays. If you are not familiar with strtok, another reference link is provided below:

http://www.cplusplus.com/reference/cstring/strtok/

I hope this helps!

answered Nov 19 '18 at 17:51

armitus

845

If you would like to continue using a scanf approach, I would recommend the negated scanset %[^] format specifier.

scanf("%[^n]", pointerToCharArray)

This should read in any number of characters up to but not including the character specified (which, in our case, is a newline). If you'd like to discard the newline, read it in as follows:

scanf("%[^n]n", pointerToCharArray)

A link to the reference page can be found below. The negated scanset specifier is included in the list:

http://www.cplusplus.com/reference/cstdio/scanf/

From this point, it is a simple matter to use strtok() to tokenize the output of the scanf into number arrays. If you are not familiar with strtok, another reference link is provided below:

http://www.cplusplus.com/reference/cstring/strtok/

I hope this helps!

answered Nov 19 '18 at 17:51

armitus

845

answered Nov 19 '18 at 17:51

armitus

845

answered Nov 19 '18 at 17:51

armitus

845

answered Nov 19 '18 at 17:51

armitus

845

scanf("%[^n]n", pointerToCharArray) not only consumes a following newline, it waits for additional input consuming all other white-spaces. scanf() then does not return until following non-whitespace detected or EOF.
– chux
Nov 19 '18 at 18:57

As OP is concerned about "I DON'T KNOW how many number ", using "%[^n]" without a width limit is fragile code.
– chux
Nov 19 '18 at 18:58

Thank you very much!
– user7331489
Nov 22 '18 at 8:43

add a comment |

scanf("%[^n]n", pointerToCharArray) not only consumes a following newline, it waits for additional input consuming all other white-spaces. scanf() then does not return until following non-whitespace detected or EOF.
– chux
Nov 19 '18 at 18:57

As OP is concerned about "I DON'T KNOW how many number ", using "%[^n]" without a width limit is fragile code.
– chux
Nov 19 '18 at 18:58

Thank you very much!
– user7331489
Nov 22 '18 at 8:43

scanf("%[^n]n", pointerToCharArray) not only consumes a following newline, it waits for additional input consuming all other white-spaces. scanf() then does not return until following non-whitespace detected or EOF.
– chux
Nov 19 '18 at 18:57

As OP is concerned about "I DON'T KNOW how many number ", using "%[^n]" without a width limit is fragile code.
– chux
Nov 19 '18 at 18:58

Thank you very much!
– user7331489
Nov 22 '18 at 8:43

add a comment |

Let us use the non standard getline, a common library extension. @Abbott to read the entire line into allocated memory.

If an array is needed and its size determined at run-time, use a variable length array available in C99 and optionally in C11.

Make a function to count and maybe save the numbers. Use strto...() functions for robust parsing.

size_t count_longs(char *line, long *dest) {

  size_t count = 0;

  char *s = line;

  for (;;) {

    char *endptr;

    long num = strtol(s, &endptr, 10);

    if (s == endptr) {

      break;  /// no conversion

    }

    s = endptr;

    if (dest) {

      dest[count] = num;

    }

    count++;

  }

  return count;

}

Sample code snippet

  char *line = NULL;

  size_t len = 0;

  ssize_t nread = getline(&line, &len, stdin);  // read the whole line

  if (nread != -1) {

    // successful read



    long a[count_longs(line, NULL)];  // determine array size and use a 

    count_longs(line, a);             // 2nd pass: assign array elements. 



    nread = getline(&line, &len, stdin);

    if (nread != -1) {

      // successful read



      long b[count_longs(line, NULL)];

      count_longs(line, b);



      // Do something with a and b

    }

  }

  free(line);

  len = 0;

edited Nov 19 '18 at 18:19

answered Nov 19 '18 at 18:05

chux

80.8k870148

add a comment |

Let us use the non standard getline, a common library extension. @Abbott to read the entire line into allocated memory.

If an array is needed and its size determined at run-time, use a variable length array available in C99 and optionally in C11.

Make a function to count and maybe save the numbers. Use strto...() functions for robust parsing.

size_t count_longs(char *line, long *dest) {

  size_t count = 0;

  char *s = line;

  for (;;) {

    char *endptr;

    long num = strtol(s, &endptr, 10);

    if (s == endptr) {

      break;  /// no conversion

    }

    s = endptr;

    if (dest) {

      dest[count] = num;

    }

    count++;

  }

  return count;

}

Sample code snippet

  char *line = NULL;

  size_t len = 0;

  ssize_t nread = getline(&line, &len, stdin);  // read the whole line

  if (nread != -1) {

    // successful read



    long a[count_longs(line, NULL)];  // determine array size and use a 

    count_longs(line, a);             // 2nd pass: assign array elements. 



    nread = getline(&line, &len, stdin);

    if (nread != -1) {

      // successful read



      long b[count_longs(line, NULL)];

      count_longs(line, b);



      // Do something with a and b

    }

  }

  free(line);

  len = 0;

edited Nov 19 '18 at 18:19

answered Nov 19 '18 at 18:05

chux

80.8k870148

add a comment |

Let us use the non standard getline, a common library extension. @Abbott to read the entire line into allocated memory.

If an array is needed and its size determined at run-time, use a variable length array available in C99 and optionally in C11.

Make a function to count and maybe save the numbers. Use strto...() functions for robust parsing.

size_t count_longs(char *line, long *dest) {

  size_t count = 0;

  char *s = line;

  for (;;) {

    char *endptr;

    long num = strtol(s, &endptr, 10);

    if (s == endptr) {

      break;  /// no conversion

    }

    s = endptr;

    if (dest) {

      dest[count] = num;

    }

    count++;

  }

  return count;

}

Sample code snippet

  char *line = NULL;

  size_t len = 0;

  ssize_t nread = getline(&line, &len, stdin);  // read the whole line

  if (nread != -1) {

    // successful read



    long a[count_longs(line, NULL)];  // determine array size and use a 

    count_longs(line, a);             // 2nd pass: assign array elements. 



    nread = getline(&line, &len, stdin);

    if (nread != -1) {

      // successful read



      long b[count_longs(line, NULL)];

      count_longs(line, b);



      // Do something with a and b

    }

  }

  free(line);

  len = 0;

edited Nov 19 '18 at 18:19

answered Nov 19 '18 at 18:05

chux

80.8k870148

Let us use the non standard getline, a common library extension. @Abbott to read the entire line into allocated memory.

If an array is needed and its size determined at run-time, use a variable length array available in C99 and optionally in C11.

Make a function to count and maybe save the numbers. Use strto...() functions for robust parsing.

size_t count_longs(char *line, long *dest) {

  size_t count = 0;

  char *s = line;

  for (;;) {

    char *endptr;

    long num = strtol(s, &endptr, 10);

    if (s == endptr) {

      break;  /// no conversion

    }

    s = endptr;

    if (dest) {

      dest[count] = num;

    }

    count++;

  }

  return count;

}

Sample code snippet

  char *line = NULL;

  size_t len = 0;

  ssize_t nread = getline(&line, &len, stdin);  // read the whole line

  if (nread != -1) {

    // successful read



    long a[count_longs(line, NULL)];  // determine array size and use a 

    count_longs(line, a);             // 2nd pass: assign array elements. 



    nread = getline(&line, &len, stdin);

    if (nread != -1) {

      // successful read



      long b[count_longs(line, NULL)];

      count_longs(line, b);



      // Do something with a and b

    }

  }

  free(line);

  len = 0;

edited Nov 19 '18 at 18:19

answered Nov 19 '18 at 18:05

chux

80.8k870148

edited Nov 19 '18 at 18:19

answered Nov 19 '18 at 18:05

chux

80.8k870148

answered Nov 19 '18 at 18:05

chux

80.8k870148

answered Nov 19 '18 at 18:05

chux

80.8k870148

add a comment |

int c = 0;

int *arr = NULL;

int *extArr = NULL;

int i = 0;

for (i = 0; scanf("%d", &c) == 1; i++)

{

    extArr = (int*)realloc(arr, (i+1) * sizeof(int)); 

    arr = extArr;

    arr[i] = c; 



}

free(arr);

edited Nov 19 '18 at 19:08

answered Nov 19 '18 at 18:37

ats

694

I do see a functional problem. scanf("%d", &c) will read through an end-of-line like any other white-space. I'd expect OP's two lines of integers to end up in one arr here.
– chux
Nov 19 '18 at 19:02

The idea was to use my codesnippet as a standalone function & then just call it n times for the number of array's required. i admittedly omitted those details from my answer.
– ats
Nov 19 '18 at 19:06

Trouble is we do not know n prior to reading the line.
– chux
Nov 19 '18 at 19:08

Maybe i should've phrased that better. I did not mean n as in array length, but n as in number of array's read. Since the codesnippet can easily be packed inside a function and used as a sort of scanf-overload for array's
– ats
Nov 19 '18 at 19:10

add a comment |

int c = 0;

int *arr = NULL;

int *extArr = NULL;

int i = 0;

for (i = 0; scanf("%d", &c) == 1; i++)

{

    extArr = (int*)realloc(arr, (i+1) * sizeof(int)); 

    arr = extArr;

    arr[i] = c; 



}

free(arr);

edited Nov 19 '18 at 19:08

answered Nov 19 '18 at 18:37

ats

694

I do see a functional problem. scanf("%d", &c) will read through an end-of-line like any other white-space. I'd expect OP's two lines of integers to end up in one arr here.
– chux
Nov 19 '18 at 19:02

The idea was to use my codesnippet as a standalone function & then just call it n times for the number of array's required. i admittedly omitted those details from my answer.
– ats
Nov 19 '18 at 19:06

Trouble is we do not know n prior to reading the line.
– chux
Nov 19 '18 at 19:08

Maybe i should've phrased that better. I did not mean n as in array length, but n as in number of array's read. Since the codesnippet can easily be packed inside a function and used as a sort of scanf-overload for array's
– ats
Nov 19 '18 at 19:10

add a comment |

int c = 0;

int *arr = NULL;

int *extArr = NULL;

int i = 0;

for (i = 0; scanf("%d", &c) == 1; i++)

{

    extArr = (int*)realloc(arr, (i+1) * sizeof(int)); 

    arr = extArr;

    arr[i] = c; 



}

free(arr);

edited Nov 19 '18 at 19:08

answered Nov 19 '18 at 18:37

ats

694

int c = 0;

int *arr = NULL;

int *extArr = NULL;

int i = 0;

for (i = 0; scanf("%d", &c) == 1; i++)

{

    extArr = (int*)realloc(arr, (i+1) * sizeof(int)); 

    arr = extArr;

    arr[i] = c; 



}

free(arr);

edited Nov 19 '18 at 19:08

answered Nov 19 '18 at 18:37

ats

694

edited Nov 19 '18 at 19:08

answered Nov 19 '18 at 18:37

ats

694

answered Nov 19 '18 at 18:37

ats

694

answered Nov 19 '18 at 18:37

ats

694

I do see a functional problem. scanf("%d", &c) will read through an end-of-line like any other white-space. I'd expect OP's two lines of integers to end up in one arr here.
– chux
Nov 19 '18 at 19:02

The idea was to use my codesnippet as a standalone function & then just call it n times for the number of array's required. i admittedly omitted those details from my answer.
– ats
Nov 19 '18 at 19:06

Trouble is we do not know n prior to reading the line.
– chux
Nov 19 '18 at 19:08

Maybe i should've phrased that better. I did not mean n as in array length, but n as in number of array's read. Since the codesnippet can easily be packed inside a function and used as a sort of scanf-overload for array's
– ats
Nov 19 '18 at 19:10

add a comment |

I do see a functional problem. scanf("%d", &c) will read through an end-of-line like any other white-space. I'd expect OP's two lines of integers to end up in one arr here.
– chux
Nov 19 '18 at 19:02

The idea was to use my codesnippet as a standalone function & then just call it n times for the number of array's required. i admittedly omitted those details from my answer.
– ats
Nov 19 '18 at 19:06

Trouble is we do not know n prior to reading the line.
– chux
Nov 19 '18 at 19:08

Maybe i should've phrased that better. I did not mean n as in array length, but n as in number of array's read. Since the codesnippet can easily be packed inside a function and used as a sort of scanf-overload for array's
– ats
Nov 19 '18 at 19:10

I do see a functional problem. scanf("%d", &c) will read through an end-of-line like any other white-space. I'd expect OP's two lines of integers to end up in one arr here.
– chux
Nov 19 '18 at 19:02

The idea was to use my codesnippet as a standalone function & then just call it n times for the number of array's required. i admittedly omitted those details from my answer.
– ats
Nov 19 '18 at 19:06

Trouble is we do not know n prior to reading the line.
– chux
Nov 19 '18 at 19:08

Maybe i should've phrased that better. I did not mean n as in array length, but n as in number of array's read. Since the codesnippet can easily be packed inside a function and used as a sort of scanf-overload for array's
– ats
Nov 19 '18 at 19:10

add a comment |

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