question regarding String.intern()

[amitabh mehra]

From java specification (3.10.5) String literals :

package testPackage;
class Test {
public static void main(String[] args) {
String hello = "Hello", lo = "lo";
System.out.print((hello == "Hello") + " ");
System.out.print((Other.hello == hello) + " ");
System.out.print((other.Other.hello == hello) + " ");
System.out.print((hello == ("Hel"+"lo")) + " ");
System.out.print((hello == ("Hel"+lo)) + " ");
System.out.println(hello == ("Hel"+lo).intern());
}
}
class Other { static String hello = "Hello"; }

and the compilation unit:

package other;
public class Other { static String hello = "Hello"; }

produces the output:

true true true true false true

This example illustrates six points:

* 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 at run time are newly created and therefore distinct.
* The result of explicitly interning a computed string is the same string as any pre-existing literal string with the same contents.

Also javadoc says:

"Returns a canonical representation for the string object.

A pool of strings, initially empty, is maintained privately by the class String.
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.

It follows that for any two strings s and t, s.intern() == t.intern() is true if and only if s.equals(t) is true.

All literal strings and string-valued constant expressions are interned."

Is the pool that is being talked about here, different from the string literal pool?
If no, then the 5th println in above mentioned example should have been true. Please explain.

Thanks

[Jesper de Jong]

You're talking about this line (which produces the 5th answer "false"):

System.out.print((hello == ("Hel"+lo)) + " ");

Note the rules:

* Strings computed by constant expressions (§15.28) are computed at compile time and then treated as if they were literals.
* Strings computed at run time are newly created and therefore distinct.

The expression "Hel"+lo is computed at runtime, and therefore creates a new String object that is distinct from the String object that hello refers to.

[amitabh mehra]

I am a bit confused about the literal pool.

Point me if I am wrong:
To cut down the number of String objects created in the JVM, the String class keeps a pool of strings. So each time a new string is generated, this pool is checked first and if it exists in the pool, a reference to it is returned. else, new string object is created and also added to pool.
Now, when I say "Hel" + lo, which finally results in "Hello", and a new object is returned (since its at runtime), does it mean that pool is now having 2 objects with same content :"Hello"?

[amitabh mehra]

Ranchers, can someone clear the doubt?

[Rob Spoor]

The string pool has at most one version of each string constant. That's why String.intern() will return the exact same object references:

String s1 = "Hello";
String s2 = new String("Hello");
String s3 = new String("Hel") + "lo";
System.out.println(s1 == s2); // false
System.out.println(s1 == s3); // false
System.out.println(s2 == s3); // false
System.out.println(s1.intern() == s2.intern()); // true
System.out.println(s1.intern() == s3.intern()); // true
System.out.println(s2.intern() == s3.intern()); // true

[amitabh mehra]

Rob Prime wrote:The string pool has at most one version of each string constant.

Rob, Thats exactly my doubt. When I said String s1 = "hello", an entry for hello has been added into the literal pool. Now when I created String s2 = new String("hello"), what will happen?

[Rob Spoor]

It will create a new object on the heap, but not in the string pool.

From the API of String.intern():

Returns a canonical representation for the string object.

A pool of strings, initially empty, is maintained privately by the class String.

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.

It follows that for any two strings s and t, s.intern() == t.intern() is true if and only if s.equals(t) is true.

All literal strings and string-valued constant expressions are interned. String literals are defined in §3.10.5 of the Java Language Specification.

Using string literals will add those to the string pool automatically, and string literals are also retrieved from the string pool.

So with "String s2 = new String("hello")", it will fetch "hello" from the string pool, and use its contents to create a new object.