s2 and s3 are reference to two different String objects.  Both objects
have the same value, but they are two different objects.  Perhaps,
you meant to write

        System.out.println(s2.intern() == s3);

The intern() method returns a reference to the String object in the pool
whose value is the same as the String on which intern() is invoked.  In
this case, s2.inter() returns a reference to the same object that s3 refers
to because string literals, such as "arit", are automatically interned.

Looking up a String in the internal interned table is a relatively
expensive operation. It's very much faster just to return another object.

Tom Hawtin

Are you just comparing the String values?  Why aren't you just using
.equals?

String s1 = new String("amit"); //Create NEW String object for "amit"
System.out.println(s1.replace('m','r'));
System.out.println(s1);
String s3 = "arit"; // "arit" not found in String pool - hence create
new obj
String s4 = "arit"; // same obj as the one referenced by s3
String s2 = s1.replace('m','r'); // NEW String object
System.out.println(s2 == s3);// Hence s2 and s3 are 2 different objects
System.out.println(s3 == s4);

cheers
attitudenine

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Look a bit more closely at the String pool rules, in, for example, the
API documentation for String's intern() method. Creating a String with
the same content as an intern String does not, in general, result in a
reference to the intern String. There are two ways of getting a
reference to a String in the intern pool:

1. The String reference is the result of evaluating a constant
expression of type String.

2. The String reference is the result of calling a String object's
intern() method.

s3 and s4 both refer to constant expressions of type String, so they
point to the "arit" in the pool.

Each of s1 and s2 contains neither the result of evaluating a String
constant expression, nor the result of an intern call. Neither should
point to the pool.

Patricia