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?










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
    yesterday










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






  • 8




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










  • 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
    yesterday















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?










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
    yesterday










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






  • 8




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










  • 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
    yesterday













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?










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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









JJJ

29k147591




29k147591










asked yesterday









Rahul Dev

1589




1589








  • 2




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










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






  • 8




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










  • 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
    yesterday














  • 2




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










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






  • 8




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










  • 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
    yesterday








2




2




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




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












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




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




8




8




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




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












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
yesterday




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
yesterday












1 Answer
1






active

oldest

votes

















up vote
16
down vote



accepted










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



















  • 4




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






  • 2




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






  • 1




    @Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
    – Sarief
    yesterday






  • 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
    yesterday








  • 2




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











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1 Answer
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1 Answer
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active

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up vote
16
down vote



accepted










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



















  • 4




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






  • 2




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






  • 1




    @Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
    – Sarief
    yesterday






  • 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
    yesterday








  • 2




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















up vote
16
down vote



accepted










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



















  • 4




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






  • 2




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






  • 1




    @Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
    – Sarief
    yesterday






  • 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
    yesterday








  • 2




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













up vote
16
down vote



accepted







up vote
16
down vote



accepted






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 yesterday

























answered yesterday









Eran

272k35431514




272k35431514








  • 4




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






  • 2




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






  • 1




    @Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
    – Sarief
    yesterday






  • 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
    yesterday








  • 2




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














  • 4




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






  • 2




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






  • 1




    @Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
    – Sarief
    yesterday






  • 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
    yesterday








  • 2




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








4




4




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




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




2




2




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




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




1




1




@Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
– Sarief
yesterday




@Eran Let me rephrase, even if it had in contract that it creates a new object and it did create a new object, it does not mean that it will return a new object. Strings are contained in Strings pool and are reused mostly from there
– Sarief
yesterday




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
yesterday






@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
yesterday






2




2




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




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


















 

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