Strange behavior with string interning in Java

Question

There is code as following:

String s = new String("1");
s.intern();
String s2 = "1";
System.out.println(s == s2);

String s3 = new String("1")+new String("1");
s3.intern();
String s4 = "11";
System.out.println(s3 == s4);

Output of the code above is:

false
true

I know that s and s2 are different objects, so the result evaluates to false, but the second result evaluates to true. Can anyone tell me the difference?

Note that calling s3.intern() after initializing s4 changes the output to false. This seems to indicate that the literal "11" is only retrieved from the pool as the line is executed, which isn't how I understood string literal interning to work. — shmosel
– shmosel, Commented Mar 16, 2017 at 4:31
@RobbyCornelissen That's essentially the same observation I made. You just moved the literal higher up. — shmosel
– shmosel, Commented Mar 16, 2017 at 4:59
@Maybe_Factor It's not an answer at all, it's just a link to an article. — shmosel
– shmosel, Commented Mar 16, 2017 at 5:26
For users that do not have enough reputation to see the graveyard of deleted answers to this question – if you're planning to write an answer that a) outlines the fundamentals of string interning in Java, or b) suggests that the result of String.intern() should be assigned back to the variable: don't bother. The real question is why the behavior in the two cases (s == s2 vs s3 == s4) is different. — Robby Cornelissen
– Robby Cornelissen, Commented Mar 16, 2017 at 5:34

Community · Accepted Answer · 2017-05-23 12:02:04Z

23

Here's what's happening:

Example 1

String s1 = new String("1"); 
s1.intern();
String s2 = "1";

The string literal "1" (passed into the String constructor) is interned at address A.
String s1 is created at address B because it is not a literal or constant expression.
The call to intern() has no effect. String "1" is already interned, and the result of the operation is not assigned back to s1.
String s2 with value "1" is retrieved from the string pool, so points to address A.

Result: Strings s1 and s2 point to different addresses.

Example 2

String s3 = new String("1") + new String("1");
s3.intern();
String s4 = "11";

String s3 is created at address C.
The call to intern() adds the string with value "11" at address C to the string pool.
String s4 with value "11" is retrieved from the string pool, so points to address C.

Result: Strings s3 and s4 point to the same address.

Summary

String "1" is interned before the call to intern() is made, by virtue of its presence in the s1 = new String("1") constructor call.

Changing that constructor call to s1 = new String(new char[]{'1'}) will make the comparison of s1 == s2 evaluate to true because both will now refer to the string that was explicitly interned by calling s1.intern().

(I used the code from this answer to get information about the strings' memory locations.)

edited May 23, 2017 at 12:02

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answered Mar 16, 2017 at 6:55

Robby Cornelissen

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13 Comments

Roland Illig Over a year ago

Ah, that makes sense now, but it requires very careful reading. Note that the observed behavior changes if any other code has already put "11" into the interned string pool before.

Robby Cornelissen Over a year ago

@RolandIllig Indeed, I noted that in the comments to the question.

shmosel Over a year ago

The steps in your breakdown can be deduced from the output, as I commented earlier. I'm much more interested in seeing where it's documented, if indeed it is.

Robby Cornelissen Over a year ago

@shmosel What would you like to see documented? That the mere use of a string literal (without assigning it) is enough to have it interned in the string pool?

shmosel Over a year ago

I meant "used" the same way you did. My point is I would have expected "a" to be interned long before that statement executes, possibly at class load. I might have been entirely wrong, but it would be nice to see a source one way or another.

|

chengpohi · Accepted Answer · 2017-03-16 07:22:01Z

12

For the scenario 1:

String s = new String("1");
s.intern();
String s2 = "1";
System.out.println(s == s2);

with bytecode:

   0: new           #2                  // class java/lang/String
   3: dup
   4: ldc           #3                  // String 1
   6: invokespecial #4                  // Method java/lang/String."<init>":(Ljava/lang/String;)V
   9: astore_1
  10: aload_1
  11: invokevirtual #5                  // Method java/lang/String.intern:()Ljava/lang/String;
  14: pop
  15: ldc           #3                  // String 1

for String s = new String("1"); it will create a new String object, it will have a new address with "1" that it is already in String Pool:

ldc #3 // String 1

and for s2, as the bytecode:

15: ldc #3 // String 1

s2 is pointing to String Pool variable: "1", so s and s2 have the different address and result is false.

For the scenario 2:

String s3 = new String("1")+new String("1");
s3.intern();
String s4 = "11";
System.out.println(s3 == s4);

with bytecode:

   0: new           #2                  // class java/lang/StringBuilder
   3: dup
   4: invokespecial #3                  // Method java/lang/StringBuilder."<init>":()V
   7: astore_1
   8: aload_1
   9: ldc           #4                  // String 1
  11: invokevirtual #5                  // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
  14: ldc           #4                  // String 1
  16: invokevirtual #5                  // Method java/lang/StringBuilder.append:(Ljava/lang/String;)Ljava/lang/StringBuilder;
  19: invokevirtual #6                  // Method java/lang/StringBuilder.toString:()Ljava/lang/String;
  22: astore_2
  23: aload_2
  24: invokevirtual #7                  // Method java/lang/String.intern:()Ljava/lang/String;
  27: astore_3
  28: ldc           #8                  // String 11

As the bytecode, you can see new String("1")+new String("1"); is created by using StringBuilder

new #2 // class java/lang/StringBuilder

it's totally a new Object without String Pool variable.

and after s3.intern(), this method will add current s3 to the Memory String Pool and 8: aload_1.

and s4 is trying to load from

ldc #8 // String 11

so s3 and s4 address should equal and result is true.

answered Mar 16, 2017 at 7:22

chengpohi

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5 Comments

Shyam Baitmangalkar Over a year ago

Perfect!!!! Your bytecode trace hits the spot and explains all of it!!

Shyam Baitmangalkar Over a year ago

Just one question here. What is the way to be followed to get a bytecode trace like this?

Shyam Baitmangalkar Over a year ago

I mean, are you using anything other than javap ?

chengpohi Over a year ago

@ShyamBaitmangalkar, Yes I am using javap for see bytecode.

Jörn Horstmann Over a year ago

How does "this method will add current s3 to the Memory String Pool" work? The String "11" is already in the constant pool, due to being used as a literal constant. You can see this with the javap -v.

Yu Jiaao · Accepted Answer · 2017-03-17 01:46:25Z

0

Just for someone who use groovy, the addition info is: the behavior is different

answered Mar 17, 2017 at 1:46

Yu Jiaao

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Comments

Maurice Perry · Accepted Answer · 2017-03-16 07:07:18Z

-3

s.intern() doesn't change the string s. You should have written:

    s = s.intern();

answered Mar 16, 2017 at 7:07

Maurice Perry

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5 Comments

Robby Cornelissen Over a year ago

So why does it seem to work for s3.intern()?. Please read my comment on the original question.

Maurice Perry Over a year ago

@ShyamBaitmangalkar yes, it does. @Robby I guess the compiler evaluates the expression. What I've noticed, is that with s = s.intern() and s3 = s3.intern(), I obtain true and true.

shmosel Over a year ago

But you don't need to reassign s3 to get true, as in OP's example.

Maurice Perry Over a year ago

@shmosel no, but the documentation says "Returns a canonical representation for the string object.", so the correct way to use .intern(), is to use the returned value (and when done this way, it behaves as expected). That said, I don't know why the original program behaves the way it does.

shmosel Over a year ago

We know it's not the correct use. But that's not the question.

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