By reading all the questions already asked in this forum related to the topic above (see title), I thoroughly understand that finally gets always called. (except from System.exit and infinite loops). However, I would like to know if a return is called in a catch block and then another return is called from the finally block.
For example:
public static void main(String[]args) {
int a = new TestClass().absorbeTheValue();
}
int absorbeTheValue() {
try {
int a = 10/0;
if (a > 0) return 4;
} catch(Exception e) {
return 45;
} finally {
return 34;
}
}
So here the output (when the method is called) is going to be 34 in any case. It means that finally always gets run. I think though that the other "returns" are not run at all. In many posts I found the fact that finally write the content over what had been already written by the catch clause return. My understanding is that as soon as the return value in the catch clause is about to be evaluated, the control flow pass to the finally clause which having in turn another return, this time the return will be evaluated without passing control back to the catch clause. In this way the only return called at runtime will be the finally return. Do you agree with that?
A return in finally does not pass back the control to the program but returns the value and terminates the method. Can we say so?
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.
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
Source: Stackoverflow.com