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Multiple returns: Which one sets the final return value? 6 answers

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If the return in the try block is reached, it transfers control to the finally block, and the function eventually returns normally (not a throw).

If an exception occurs, but then the code reaches a return from the catch block, control is transferred to the finally block and the function eventually returns normally (not a throw).

In your example, you have a return in the finally , and so regardless of what happens, the function will return 34 , because finally has the final (if you will) word.

Although not covered in your example, this would be true even if you didn't have the catch and if an exception were thrown in the try block and not caught. By doing a return from the finally block, you suppress the exception entirely. Consider:

public class FinallyReturn {
  public static final void main(String[] args) {
    System.out.println(foo(args));
  }

  private static int foo(String[] args) {
    try {
      int n = Integer.parseInt(args[0]);
      return n;
    }
    finally {
      return 42;
    }
  }
}

If you run that without supplying any arguments:

$ java FinallyReturn

...the code in foo throws an ArrayIndexOutOfBoundsException . But because the finally block does a return , that exception gets suppressed.

This is one reason why it's best to avoid using return in finally .

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Here is some code that show how it works.

class Test
{
    public static void main(String args[]) 
    { 
        System.out.println(Test.test()); 
    }

    public static String test()
    {
        try {
            System.out.println("try");
            throw new Exception();
        } catch(Exception e) {
            System.out.println("catch");
            return "return"; 
        } finally {  
            System.out.println("finally");
            return "return in finally"; 
        }
    }
}

The results is:

try
catch
finally
return in finally

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