Mixing
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
edited 8 hours ago
JJJ
29k147591
asked 9 hours ago
Rahul Dev
1568
add a comment |
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
edited 8 hours ago
JJJ
29k147591
asked 9 hours ago
Rahul Dev
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
add a comment |
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
edited 8 hours ago
JJJ
29k147591
asked 9 hours ago
Rahul Dev
1568
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 8 hours ago
JJJ
29k147591
asked 9 hours ago
Rahul Dev
1568
edited 8 hours ago
JJJ
29k147591
asked 9 hours ago
Rahul Dev
1568
edited 8 hours ago
JJJ
29k147591
edited 8 hours ago
JJJ
29k147591
edited 8 hours ago
JJJ
29k147591
29k147591
asked 9 hours ago
Rahul Dev
1568
asked 9 hours ago
Rahul Dev
1568
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
add a comment |
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
add a comment |
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
}
....
}
answered 9 hours ago
Eran
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 == ywould 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 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
9 hours ago
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
9 hours ago
|
show 8 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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
}
....
}
answered 9 hours ago
Eran
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 == ywould 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 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
9 hours ago
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
9 hours ago
|
show 8 more comments
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
}
....
}
answered 9 hours ago
Eran
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 == ywould 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 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
9 hours ago
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
9 hours ago
|
show 8 more comments
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
}
....
}
answered 9 hours ago
Eran
272k35430514
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 9 hours ago
Eran
272k35430514
edited 9 hours ago
answered 9 hours ago
Eran
272k35430514
answered 9 hours ago
Eran
272k35430514
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 == ywould 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 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
9 hours ago
2
@nits.kk this should not apply for immutable objects, which Strings are
– Sarief
9 hours ago
|
show 8 more comments
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 == ywould 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 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
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
|
show 8 more comments
draft saved
draft discarded
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53370200%2fwhen-does-the-touppercase-method-create-a-new-object%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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