When does the toUpperCase() method create a new object?











up vote
11
down vote

favorite
1












public class Child{



    public static void main(String args){

        String x = new String("ABC");

        String y = x.toUpperCase();



        System.out.println(x == y);

    }

}


Output: true



So does toUpperCase() always create a new object?






java string object







share|improve this question




















  • 2




    I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
    – Peter Lawrey
    9 hours ago










  • Note: new String(...) doesn't change the answer.
    – Peter Lawrey
    9 hours ago






  • 8




    String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
    – Sarief
    9 hours ago










  • edit: someone pointed that OP used new String("ABC") to point to fact that it's not interned. I don't see how interning or not interning makes difference for toUpperCase(Locale) method
    – Sarief
    9 hours ago















up vote
11
down vote

favorite
1












public class Child{



    public static void main(String args){

        String x = new String("ABC");

        String y = x.toUpperCase();



        System.out.println(x == y);

    }

}


Output: true



So does toUpperCase() always create a new object?






java string object







share|improve this question




















  • 2




    I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
    – Peter Lawrey
    9 hours ago










  • Note: new String(...) doesn't change the answer.
    – Peter Lawrey
    9 hours ago






  • 8




    String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
    – Sarief
    9 hours ago










  • edit: someone pointed that OP used new String("ABC") to point to fact that it's not interned. I don't see how interning or not interning makes difference for toUpperCase(Locale) method
    – Sarief
    9 hours ago













up vote
11
down vote

favorite
1









up vote
11
down vote

favorite
1






1





public class Child{



    public static void main(String args){

        String x = new String("ABC");

        String y = x.toUpperCase();



        System.out.println(x == y);

    }

}


Output: true



So does toUpperCase() always create a new object?






java string object







share|improve this question















public class Child{



    public static void main(String args){

        String x = new String("ABC");

        String y = x.toUpperCase();



        System.out.println(x == y);

    }

}


Output: true



So does toUpperCase() always create a new object?






java string object




java string object






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 hours ago









JJJ

29k147591




29k147591










asked 9 hours ago









Rahul Dev

1568




1568








  • 2




    I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
    – Peter Lawrey
    9 hours ago










  • Note: new String(...) doesn't change the answer.
    – Peter Lawrey
    9 hours ago






  • 8




    String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
    – Sarief
    9 hours ago










  • edit: someone pointed that OP used new String("ABC") to point to fact that it's not interned. I don't see how interning or not interning makes difference for toUpperCase(Locale) method
    – Sarief
    9 hours ago














  • 2




    I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
    – Peter Lawrey
    9 hours ago










  • Note: new String(...) doesn't change the answer.
    – Peter Lawrey
    9 hours ago






  • 8




    String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
    – Sarief
    9 hours ago










  • edit: someone pointed that OP used new String("ABC") to point to fact that it's not interned. I don't see how interning or not interning makes difference for toUpperCase(Locale) method
    – Sarief
    9 hours ago








2




2




I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
9 hours ago




I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
9 hours ago












Note: new String(...) doesn't change the answer.
– Peter Lawrey
9 hours ago




Note: new String(...) doesn't change the answer.
– Peter Lawrey
9 hours ago




8




8




String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
9 hours ago




String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
9 hours ago












edit: someone pointed that OP used new String("ABC") to point to fact that it's not interned. I don't see how interning or not interning makes difference for toUpperCase(Locale) method
– Sarief
9 hours ago




edit: someone pointed that OP used new String("ABC") to point to fact that it's not interned. I don't see how interning or not interning makes difference for toUpperCase(Locale) method
– Sarief
9 hours ago












1 Answer
1






active

oldest

votes

















up vote
16
down vote













toUpperCase() calls toUpperCase(Locale.getDefault()), which creates a new String object only if it has to. If the input String is already in upper case, it returns the input String.



This seems to be an implementation detail, though. I didn't find it mentioned in the Javadoc.



Here's an implementation:



public String toUpperCase(Locale locale) {

    if (locale == null) {

        throw new NullPointerException();

    }



    int firstLower;

    final int len = value.length;



    /* Now check if there are any characters that need to be changed. */

    scan: {

        for (firstLower = 0 ; firstLower < len; ) {

            int c = (int)value[firstLower];

            int srcCount;

            if ((c >= Character.MIN_HIGH_SURROGATE)

                    && (c <= Character.MAX_HIGH_SURROGATE)) {

                c = codePointAt(firstLower);

                srcCount = Character.charCount(c);

            } else {

                srcCount = 1;

            }

            int upperCaseChar = Character.toUpperCaseEx(c);

            if ((upperCaseChar == Character.ERROR)

                    || (c != upperCaseChar)) {

                break scan;

            }

            firstLower += srcCount;

        }

        return this; // <-- the original String is returned

    }

    ....

}





share|improve this answer























  • Exactly, nothing seems to be documented. which is why I was confused.
    – Rahul Dev
    9 hours ago






  • 3




    @Sarief if it created a new object (and returned that new object) x == y would definitely return false.
    – Eran
    9 hours ago






  • 2




    @Sarief it somehow got to this list, which tends to result in high traffic.
    – Eran
    9 hours ago






  • 1




    @Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
    – nits.kk
    9 hours ago








  • 2




    @nits.kk this should not apply for immutable objects, which Strings are
    – Sarief
    9 hours ago











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up vote
16
down vote













toUpperCase() calls toUpperCase(Locale.getDefault()), which creates a new String object only if it has to. If the input String is already in upper case, it returns the input String.



This seems to be an implementation detail, though. I didn't find it mentioned in the Javadoc.



Here's an implementation:



public String toUpperCase(Locale locale) {

    if (locale == null) {

        throw new NullPointerException();

    }



    int firstLower;

    final int len = value.length;



    /* Now check if there are any characters that need to be changed. */

    scan: {

        for (firstLower = 0 ; firstLower < len; ) {

            int c = (int)value[firstLower];

            int srcCount;

            if ((c >= Character.MIN_HIGH_SURROGATE)

                    && (c <= Character.MAX_HIGH_SURROGATE)) {

                c = codePointAt(firstLower);

                srcCount = Character.charCount(c);

            } else {

                srcCount = 1;

            }

            int upperCaseChar = Character.toUpperCaseEx(c);

            if ((upperCaseChar == Character.ERROR)

                    || (c != upperCaseChar)) {

                break scan;

            }

            firstLower += srcCount;

        }

        return this; // <-- the original String is returned

    }

    ....

}





share|improve this answer























  • Exactly, nothing seems to be documented. which is why I was confused.
    – Rahul Dev
    9 hours ago






  • 3




    @Sarief if it created a new object (and returned that new object) x == y would definitely return false.
    – Eran
    9 hours ago






  • 2




    @Sarief it somehow got to this list, which tends to result in high traffic.
    – Eran
    9 hours ago






  • 1




    @Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
    – nits.kk
    9 hours ago








  • 2




    @nits.kk this should not apply for immutable objects, which Strings are
    – Sarief
    9 hours ago















up vote
16
down vote













toUpperCase() calls toUpperCase(Locale.getDefault()), which creates a new String object only if it has to. If the input String is already in upper case, it returns the input String.



This seems to be an implementation detail, though. I didn't find it mentioned in the Javadoc.



Here's an implementation:



public String toUpperCase(Locale locale) {

    if (locale == null) {

        throw new NullPointerException();

    }



    int firstLower;

    final int len = value.length;



    /* Now check if there are any characters that need to be changed. */

    scan: {

        for (firstLower = 0 ; firstLower < len; ) {

            int c = (int)value[firstLower];

            int srcCount;

            if ((c >= Character.MIN_HIGH_SURROGATE)

                    && (c <= Character.MAX_HIGH_SURROGATE)) {

                c = codePointAt(firstLower);

                srcCount = Character.charCount(c);

            } else {

                srcCount = 1;

            }

            int upperCaseChar = Character.toUpperCaseEx(c);

            if ((upperCaseChar == Character.ERROR)

                    || (c != upperCaseChar)) {

                break scan;

            }

            firstLower += srcCount;

        }

        return this; // <-- the original String is returned

    }

    ....

}





share|improve this answer























  • Exactly, nothing seems to be documented. which is why I was confused.
    – Rahul Dev
    9 hours ago






  • 3




    @Sarief if it created a new object (and returned that new object) x == y would definitely return false.
    – Eran
    9 hours ago






  • 2




    @Sarief it somehow got to this list, which tends to result in high traffic.
    – Eran
    9 hours ago






  • 1




    @Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
    – nits.kk
    9 hours ago








  • 2




    @nits.kk this should not apply for immutable objects, which Strings are
    – Sarief
    9 hours ago













up vote
16
down vote










up vote
16
down vote









toUpperCase() calls toUpperCase(Locale.getDefault()), which creates a new String object only if it has to. If the input String is already in upper case, it returns the input String.



This seems to be an implementation detail, though. I didn't find it mentioned in the Javadoc.



Here's an implementation:



public String toUpperCase(Locale locale) {

    if (locale == null) {

        throw new NullPointerException();

    }



    int firstLower;

    final int len = value.length;



    /* Now check if there are any characters that need to be changed. */

    scan: {

        for (firstLower = 0 ; firstLower < len; ) {

            int c = (int)value[firstLower];

            int srcCount;

            if ((c >= Character.MIN_HIGH_SURROGATE)

                    && (c <= Character.MAX_HIGH_SURROGATE)) {

                c = codePointAt(firstLower);

                srcCount = Character.charCount(c);

            } else {

                srcCount = 1;

            }

            int upperCaseChar = Character.toUpperCaseEx(c);

            if ((upperCaseChar == Character.ERROR)

                    || (c != upperCaseChar)) {

                break scan;

            }

            firstLower += srcCount;

        }

        return this; // <-- the original String is returned

    }

    ....

}





share|improve this answer














toUpperCase() calls toUpperCase(Locale.getDefault()), which creates a new String object only if it has to. If the input String is already in upper case, it returns the input String.



This seems to be an implementation detail, though. I didn't find it mentioned in the Javadoc.



Here's an implementation:



public String toUpperCase(Locale locale) {

    if (locale == null) {

        throw new NullPointerException();

    }



    int firstLower;

    final int len = value.length;



    /* Now check if there are any characters that need to be changed. */

    scan: {

        for (firstLower = 0 ; firstLower < len; ) {

            int c = (int)value[firstLower];

            int srcCount;

            if ((c >= Character.MIN_HIGH_SURROGATE)

                    && (c <= Character.MAX_HIGH_SURROGATE)) {

                c = codePointAt(firstLower);

                srcCount = Character.charCount(c);

            } else {

                srcCount = 1;

            }

            int upperCaseChar = Character.toUpperCaseEx(c);

            if ((upperCaseChar == Character.ERROR)

                    || (c != upperCaseChar)) {

                break scan;

            }

            firstLower += srcCount;

        }

        return this; // <-- the original String is returned

    }

    ....

}






share|improve this answer














share|improve this answer



share|improve this answer








edited 9 hours ago

























answered 9 hours ago









Eran

272k35430514




272k35430514












  • Exactly, nothing seems to be documented. which is why I was confused.
    – Rahul Dev
    9 hours ago






  • 3




    @Sarief if it created a new object (and returned that new object) x == y would definitely return false.
    – Eran
    9 hours ago






  • 2




    @Sarief it somehow got to this list, which tends to result in high traffic.
    – Eran
    9 hours ago






  • 1




    @Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
    – nits.kk
    9 hours ago








  • 2




    @nits.kk this should not apply for immutable objects, which Strings are
    – Sarief
    9 hours ago


















  • Exactly, nothing seems to be documented. which is why I was confused.
    – Rahul Dev
    9 hours ago






  • 3




    @Sarief if it created a new object (and returned that new object) x == y would definitely return false.
    – Eran
    9 hours ago






  • 2




    @Sarief it somehow got to this list, which tends to result in high traffic.
    – Eran
    9 hours ago






  • 1




    @Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
    – nits.kk
    9 hours ago








  • 2




    @nits.kk this should not apply for immutable objects, which Strings are
    – Sarief
    9 hours ago
















Exactly, nothing seems to be documented. which is why I was confused.
– Rahul Dev
9 hours ago




Exactly, nothing seems to be documented. which is why I was confused.
– Rahul Dev
9 hours ago




3




3




@Sarief if it created a new object (and returned that new object) x == y would definitely return false.
– Eran
9 hours ago




@Sarief if it created a new object (and returned that new object) x == y would definitely return false.
– Eran
9 hours ago




2




2




@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
9 hours ago




@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
9 hours ago




1




1




@Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
– nits.kk
9 hours ago






@Sarief First of all its not silly question, as Strings are immutable so any operation performed is expected to result in a separate object. Secondly the answer as many would think is related to same string in string pool, but its not as OP has created string using new and not invoked intern() hence correctly the answer points the implementation bwing the result as why no new object was created for the case presented by the question
– nits.kk
9 hours ago






2




2




@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
9 hours ago




@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
9 hours ago


















 

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