Why does changing the returned variable in a finally block not change the return value?

Question

I have a simple Java class as shown below:

public class Test {

    private String s;

    public String foo() {
        try {
            s = "dev";
            return s;
        } 
        finally {
            s = "override variable s";
            System.out.println("Entry in finally Block");  
        }
    }

    public static void main(String[] xyz) {
        Test obj = new Test();
        System.out.println(obj.foo());
    }
}

And the output of this code is this:

Entry in finally Block
dev  

Why is s not overridden in the finally block, yet control printed output?

Answer 1

The try block completes with the execution of the return statement and the value of s at the time the return statement executes is the value returned by the method. The fact that the finally clause later changes the value of s (after the return statement completes) does not (at that point) change the return value.

Note that the above deals with changes to the value of s itself in the finally block, not to the object that s references. If s was a reference to a mutable object (which String is not) and the contents of the object were changed in the finally block, then those changes would be seen in the returned value.

The detailed rules for how all this operates can be found in Section 14.20.2 of the Java Language Specification. Note that execution of a return statement counts as an abrupt termination of the try block (the section starting "If execution of the try block completes abruptly for any other reason R...." applies). See Section 14.17 of the JLS for why a return statement is an abrupt termination of a block.

By way of further detail: if both the try block and the finally block of a try-finally statement terminate abruptly because of return statements, then the following rules from §14.20.2 apply:

If execution of the try block completes abruptly for any other reason R [besides throwing an exception], then the finally block is executed, and then there is a choice:

• If the finally block completes normally, then the try statement completes abruptly for reason R.
• If the finally block completes abruptly for reason S, then the try statement completes abruptly for reason S (and reason R is discarded).

The result is that the return statement in the finally block determines the return value of the entire try-finally statement, and the returned value from the try block is discarded. A similar thing occurs in a try-catch-finally statement if the try block throws an exception, it is caught by a catch block, and both the catch block and the finally block have return statements.

Answer 2

Because the return value is put on the stack before the call to finally.

Answer 3

If we look inside bytecode, we'll notice that JDK has made a significant optimization, and foo() method looks like:

String tmp = null;
try {
    s = "dev"
    tmp = s;
    s = "override variable s";
    return tmp;
} catch (RuntimeException e){
    s = "override variable s";
    throw e;
}

And bytecode:

0:  ldc #7;         //loading String "dev"
2:  putstatic   #8; //storing it to a static variable
5:  getstatic   #8; //loading "dev" from a static variable
8:  astore_0        //storing "dev" to a temp variable
9:  ldc #9;         //loading String "override variable s"
11: putstatic   #8; //setting a static variable
14: aload_0         //loading a temp avariable
15: areturn         //returning it
16: astore_1
17: ldc #9;         //loading String "override variable s"
19: putstatic   #8; //setting a static variable
22: aload_1
23: athrow

java preserved "dev" string from being changed before returning. In fact here is no finally block at all.

Answer 4

There are 2 things noteworthy here:

• Strings are immutable. When you set s to "override variable s", you set s to refer to the inlined String, not altering the inherent char buffer of the s object to change to "override variable s".
• You put a reference to the s on the stack to return to the calling code. Afterwards (when the finally block runs), altering the reference should not do anything for the return value already on the stack.

Answer 5

I change your code a bit to prove the point of Ted.

As you can see in the output s is indeed changed but after the return.

public class Test {

public String s;

public String foo() {

    try {
        s = "dev";
        return s;
    } finally {
        s = "override variable s";
        System.out.println("Entry in finally Block");

    }
}

public static void main(String[] xyz) {
    Test obj = new Test();
    System.out.println(obj.foo());
    System.out.println(obj.s);
}
}

Output:

Entry in finally Block 
dev 
override variable s

Answer 6

Technically speaking, the return in the try block won't be ignored if a finally block is defined, only if that finally block also includes a return.

It's a dubious design decision that was probably a mistake in retrospect (much like references being nullable/mutable by default, and, according to some, checked exceptions). In many ways this behaviour is exactly consistent with the colloquial understanding of what finally means - "no matter what happens beforehand in the try block, always run this code." Hence if you return true from a finally block, the overall effect must always to be to return s, no?

In general, this is seldom a good idiom, and you should use finally blocks liberally for cleaning up/closing resources but rarely if ever return a value from them.

Answer 7

Try this: If you want to print the override value of s.

finally {
    s = "override variable s";    
    System.out.println("Entry in finally Block");
    return s;
}