Not exactly.  String overrides Object.hashCode, so the has code for a
String doesn't say anything about object identity, but rather something
about the contents of the String.  For example, the following code:

   String a = "Hello";
   String b = new String(a);

will create two different String objects, but they will have the same
hash code.

To get some kind of clue, you should use the static method
System.identityHashCode(Object) instead.  That will return the IDENTITY
hash code for the object, which is the one that Object.hashCode would
have returned if it were not overridden.  As you said, that provides
incomplete information, specifically:

1. If the hash codes are the same, then you don't know anything.  The
objects may be the same, or they may be different.

2. If the hash codes are different, then the objects are different.

Of course, you can compare identity more easily and get a deterministic
answer using '=='.

   This is a risky test, as it's possible for two different objects to have
the same hashcode. Much simpler is to use the == operator:

<untestedCode>
public class Test{
 public static void main(String[] args){
   String str1 = "Hello World";
   System.out.println(str1 == str1.toUpperCase());
 }
}
</untestedCode>

The above should print "false", indicating that they are indeed different
objects. In fact, you can go further and see that when you call
toUpperCase(), the original object was not modified (which is normal, since
it's immutable):

<untestedCode>
public class Test{
 public static void main(String[] args){
   String str1 = "Hello World";
   System.out.println(str1);
   System.out.println(str1.toUpperCase());
   System.out.println(str1);
 }
}
</untestedCode>

The above should print out:

<output>
Hello World
HELLO WORLD
Hello World
</output>

   Thus showing that the original String has not been modified.

   - Oliver