When does the toUpperCase() method create a new object?
12
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
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
add a comment |
12
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
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
2
I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
Nov 19 '18 at 8:00
Note: new String(...) doesn't change the answer.
– Peter Lawrey
Nov 19 '18 at 8:01
8
String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
Nov 19 '18 at 8:02
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
Nov 19 '18 at 8:21
add a comment |
12
12
12
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
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
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
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
edited Nov 19 '18 at 9:05
JJJ
29.1k147591
29.1k147591
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
asked Nov 19 '18 at 7:38
Rahul DevRahul Dev
16311
16311
2
I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
Nov 19 '18 at 8:00
Note: new String(...) doesn't change the answer.
– Peter Lawrey
Nov 19 '18 at 8:01
8
String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
Nov 19 '18 at 8:02
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
Nov 19 '18 at 8:21
add a comment |
2
I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
Nov 19 '18 at 8:00
Note: new String(...) doesn't change the answer.
– Peter Lawrey
Nov 19 '18 at 8:01
8
String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
Nov 19 '18 at 8:02
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
Nov 19 '18 at 8:21
2
2
I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
Nov 19 '18 at 8:00
I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
Nov 19 '18 at 8:00
Note: new String(...) doesn't change the answer.
– Peter Lawrey
Nov 19 '18 at 8:01
Note: new String(...) doesn't change the answer.
– Peter Lawrey
Nov 19 '18 at 8:01
8
8
String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
Nov 19 '18 at 8:02
String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
Nov 19 '18 at 8:02
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
Nov 19 '18 at 8:21
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
Nov 19 '18 at 8:21
add a comment |
1 Answer
1
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oldest
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18
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
}
....
}
answered Nov 19 '18 at 7:40
EranEran
284k37463551
4
@Sarief if it created a new object (and returned that new object)x == ywould definitely return false.
– Eran
Nov 19 '18 at 8:06
2
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
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
Nov 19 '18 at 8:09
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 usingnewand not invokedintern()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
Nov 19 '18 at 8:12
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
|
show 7 more comments
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18
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
}
....
}
answered Nov 19 '18 at 7:40
EranEran
284k37463551
4
@Sarief if it created a new object (and returned that new object)x == ywould definitely return false.
– Eran
Nov 19 '18 at 8:06
2
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
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
Nov 19 '18 at 8:09
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 usingnewand not invokedintern()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
Nov 19 '18 at 8:12
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
|
show 7 more comments
18
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
}
....
}
answered Nov 19 '18 at 7:40
EranEran
284k37463551
4
@Sarief if it created a new object (and returned that new object)x == ywould definitely return false.
– Eran
Nov 19 '18 at 8:06
2
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
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
Nov 19 '18 at 8:09
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 usingnewand not invokedintern()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
Nov 19 '18 at 8:12
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
|
show 7 more comments
18
18
18
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
}
....
}
answered Nov 19 '18 at 7:40
EranEran
284k37463551
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
}
....
}
answered Nov 19 '18 at 7:40
EranEran
284k37463551
edited Nov 19 '18 at 7:43
answered Nov 19 '18 at 7:40
EranEran
284k37463551
answered Nov 19 '18 at 7:40
EranEran
284k37463551
answered Nov 19 '18 at 7:40
EranEran
284k37463551
284k37463551
4
@Sarief if it created a new object (and returned that new object)x == ywould definitely return false.
– Eran
Nov 19 '18 at 8:06
2
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
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
Nov 19 '18 at 8:09
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 usingnewand not invokedintern()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
Nov 19 '18 at 8:12
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
|
show 7 more comments
4
@Sarief if it created a new object (and returned that new object)x == ywould definitely return false.
– Eran
Nov 19 '18 at 8:06
2
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
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
Nov 19 '18 at 8:09
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 usingnewand not invokedintern()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
Nov 19 '18 at 8:12
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
4
4
@Sarief if it created a new object (and returned that new object)
x == y would definitely return false.– Eran
Nov 19 '18 at 8:06
@Sarief if it created a new object (and returned that new object)
x == y would definitely return false.– Eran
Nov 19 '18 at 8:06
2
2
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
@Sarief it somehow got to this list, which tends to result in high traffic.
– Eran
Nov 19 '18 at 8:08
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
Nov 19 '18 at 8:09
@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
Nov 19 '18 at 8:09
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
Nov 19 '18 at 8:12
@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
Nov 19 '18 at 8:12
2
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
Nov 19 '18 at 8:23
|
show 7 more comments
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2
I wouldn't rely on this behaviour but I would expect it to avoid creating a new object.
– Peter Lawrey
Nov 19 '18 at 8:00
Note: new String(...) doesn't change the answer.
– Peter Lawrey
Nov 19 '18 at 8:01
8
String x = new String("ABC"); Please don't do this. You create String object twice. just use x = "ABC";
– Sarief
Nov 19 '18 at 8:02
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
Nov 19 '18 at 8:21