Java equals or ==

Question

I know that in case of String class, you must normally use .equals() where comparing two strings, like:

String s1 = "123";
String s2 = "123";
String s3 = "1234";
s1.equals(s2); // true
s1.equals(s3); // false

but, i've tried today this code:

s1 == s2; // true, but different references so had to return false
s1 == s3; // false

so, is something changed in java? does now string use standard comparable implementation when using == ?

Please read [How do I compare strings in Java?](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) String interning is mentioned there. — PakkuDon, Apr 02 '14 at 09:15

Marko Topolnik · Answer 1 · 2014-04-02T09:21:55.270

3

The expression "123" is not a String instance creation expression; it denotes a reference to a singleton String instance representing that string value in the string pool.

Therefore the values of your variables s1 and s2 are in fact the same.

edited Apr 02 '14 at 09:21

answered Apr 02 '14 at 09:14

Marko Topolnik

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"*Therefore your references s1 and s2 are in fact the same.*" rather Therefore your references s1 and s2 are in fact referring the same object in the pool. – Subhrajyoti Majumder Apr 02 '14 at 09:18
@SubhrajyotiMajumder A reference is a value just like 3 is a value. Two occurences of `3` in code are the same value and two references to the same object are the same value. – Marko Topolnik Apr 02 '14 at 09:19
+1 You are right. It is easy to understand with it. – Subhrajyoti Majumder Apr 02 '14 at 09:21

Abimaran Kugathasan · Accepted Answer · 2014-04-02T09:20:52.127

2

s1 == s2; // true, but different references so had to return false

No, both are referring same String literal object "123" in String literal pool. Hence returned true, when you checked with ==

If you use String s = "Some" to create object, the object is referred from String Constant Pool.

If you use the new operator to create the object, then in heap String objects will be created.

edited Apr 02 '14 at 09:20

answered Apr 02 '14 at 09:15

Abimaran Kugathasan

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if there is no "Some" in String pool, than it will be created and added to the pool, right? – Seby Apr 02 '14 at 09:30
@Seby : Yes, It's correct. – Abimaran Kugathasan Apr 02 '14 at 09:32
so it's the same of new String("123").intern() ? – Seby Apr 02 '14 at 09:32
@Seby : Yes, it's same – Abimaran Kugathasan Apr 02 '14 at 09:34
@Seby : Check the example from here http://stackoverflow.com/a/1855183/515034 – Abimaran Kugathasan Apr 02 '14 at 09:37
`"Some"` will not be created upon reaching the line `String s = "Some"`, it will already have been there since the class was loaded. – Marko Topolnik Apr 02 '14 at 09:48

score 1 · Answer 3 · answered Apr 02 '14 at 09:17

1

== checks only for the object referred by the reference.

String s1 = "123";
String s2 = "123";

Here s1 and s2 pointing the same string pool object "123"

answered Apr 02 '14 at 09:17

Deepu--Java

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score 1 · Answer 4 · answered Apr 02 '14 at 09:23

1

String s1 = new String("123");
String s2 = new String("123");

s1.equals(s2); true , because this approach is not checking the values rather the objects but s1==s2; false in this case.

answered Apr 02 '14 at 09:23

mstfdz

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score 0 · Answer 5 · answered Apr 02 '14 at 09:20

If you assign a literal to a String variable, it will point to the internalized String found in the String pool. But if you instantiate a String by new String("123") you get a new one. If you like to avoid this, using only singletons from the String pool, you have to call new String("123").intern(), which puts the new instance into the pool, or if an equal String is already in the pool, returns its reference. From the JavaDoc:

When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the #equals(Object) method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.

Java equals or ==

5 Answers5