Java String problems

Question

I ran the following program,

    String firstString = "String";
    String secondString = "String";
    String thirdString = new String("String");
    System.out.println(firstString == secondString);
    System.out.println(firstString == thirdString);
    System.out.println(firstString.intern() == thirdString);
    System.out.println(firstString.intern() == thirdString.intern());
    System.out.println(firstString.intern().equals(thirdString.intern()));
    System.out.println(firstString == thirdString);

and my output was

true
false
false
true
true
false

I learnt that the jvm pools string with same content as same strings. Is that right? If thats true then why not the firstString == thirdString return false? Does jvm only pool the string only initialized with :"" and not with new operator?

Try also: firstString == thirdString.intern(), to see that the string literals are automatically internalized. — Has QUIT--Anony-Mousse
– Has QUIT--Anony-Mousse, Commented Sep 9, 2012 at 12:59

RonK · Accepted Answer · 2012-09-09 13:14:15Z

6

The pooling relates to string literals only - so firstString and secondString are actually the same object - where as in thirdString you explicitly asked for a new object to be created on the heap.

I recommend reading the section about string literals in the spec.

It provides more information on how and when strings are pooled.

Also, take note to these bullets at the end of the section:

Literal strings within the same class (§8) in the same package (§7) represent references to the same String object (§4.3.1).

Literal strings within different classes in the same package represent references to the same String object.

Literal strings within different classes in different packages likewise represent references to the same String object.

Strings computed by constant expressions (§15.28) are computed at compile time and then treated as if they were literals.

Strings computed by concatenation at run time are newly created and therefore distinct.

edited Sep 9, 2012 at 13:14

answered Sep 9, 2012 at 12:55

RonK

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1 Comment

vvekselva Over a year ago

Okay I agree to your explanations. I've created a new object as thirdObject with "String" in heap. I've interned it. So in the pool there will be two objects right? The first and the secong string will refer to the object A(for citation) and the third will refer another(B).

Simon J. Liu · Accepted Answer · 2012-09-09 13:04:34Z

2

For firstString and secondString, JVM will lookup the string pool and return the reference of "String".

For thirdString, JVM will not lookup the string pool and will only create a String object in heap.

For OneString.intern(), JVM will lookup the reference in string pool, add OneString to string pool if OneString does not exist in it, and return the reference.

answered Sep 9, 2012 at 13:04

Simon J. Liu

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4 Comments

vvekselva Over a year ago

I have a query, adding Objects (string) in pool is based on the object hash or object contents (canonical representation). Calling intern will add into the pool and return the canonical representation that means all the string should refer to the same one since all have the same canonical representation.

Simon J. Liu Over a year ago

After calling intern(), the string will be added to string pool and the String object will refer to the new string reference in string pool. The original string object will be collected by GC next time.

vvekselva Over a year ago

Is there are any way to profile this? I just want to check the contents of the intern pool? is there are any ways?

vvekselva Over a year ago

Simon I've a query. Initally before the execution of the third line the pool will contail only one string object, after the intern method called on the third string then which object(s) will be on the pool? Either the first object or the second object added to the pool by calling the intern method(in this case thirdObject) or both the objects?

amit · Accepted Answer · 2012-09-09 12:56:10Z

0

thirdString is NOT from the pool. It is not a string literal, you dynamically created it with the new operator.

secondString on the other hand - is taken from the pool (you assign a string literal to it), so the same object is assigned to both firstString and secondString.

answered Sep 9, 2012 at 12:56

amit

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Comments

Roberto Mereghetti · Accepted Answer · 2012-09-09 13:02:17Z

0

"firstString == thirdString" returns false.

The method intern "returns a canonical representation for the string object." If you assign the interned string:

thirdString=thirdString.intern();

last "firstString == thirdString" returns true

edited Sep 9, 2012 at 13:02

answered Sep 9, 2012 at 12:56

Roberto Mereghetti

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