Try-catch-finally-return clarification [duplicate]
Solution 1:
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
.
Solution 2:
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