== Operator again

[Alamu Vinai]

Can anyone explain this when objects are created(runtime,compile time)
String st0 = "JAVA";
System.out.println(st0 == "JA" + "VA"); // prints true;

String st1 = "JAVA";
String st2 = "JA";
String st3 = "VA";
System.out.println(st1 == st2 + st3); // prints false;

[Ishaan Mohan]

In first case "Ja" + "va" it is solved at compile time.
In second case
String st1 = "JAVA";
String st2 = "JA";
String st3 = "VA";
System.out.println(st1 == st2 + st3); It is computed at runtime and "Strings computed at runtime are newly created and therefore distinct.

[William Brogden]

When the compiler sees literals like
"JA" + "VA"
or 2 * 3.14159
that can be evaluated at once, it does the evaluation, so your example is simplified by the compiler to
"JAVA"
Bill

[Cindy Glass]

String Literals are created in the Constant Pool and are resolved at compile time and created at class load time. They are also "re-used".
stO points to the "JAVA" literal.
"JA" + "VA" points to the SAME literal. The References are the same.
st1 points to "JAVA".
st2 points to a different literal "JA"
st3 points to "VA"
The system does not resolve the concatenating of the two variable references in st2 and st3 until runtime, so it does not use the Constant Pool because that has already been set in stone and the system does not KNOW that an appropriate literal will exist there. So it makes another string instead.
[This message has been edited by Cindy Glass (edited February 28, 2001).]

[Cindy Glass]

I originaly said that the new string would be created on the heap. Then it bugged me ALL LAST NIGHT. So I took that statement out.
I don't believe that I can say FOR A FACT that every JVM would handle the new string the same way.