Post increment behavior when throwing an exception

Question

Came accross this weird issue yesterday. Can't seem to find any logical explanation in the java spec for the following...

public class Program {

    public static void main(String[] args) {
        int i = 0;

        try {
            bork(i++);
        } catch (Exception e) {
        }
        System.out.println(i);
    }

    private static void bork(int i) {
        throw new RuntimeException();
    }

}

One would think post-increment would not happen because bork throws the exception, but it does!

What is the explanation for this behavior?

`i++` increments `i` and evalutes to the previous value. That has to happen before the evaluated value can be passed to a method. — khelwood, Oct 18 '19 at 14:15

score 3 · Accepted Answer · answered Oct 18 '19 at 14:14

3

The i++ operation is invoked, before the method bork() is invoked. That's the reason why it's still incremented.

So:

i++ is invoked.
As a result it returns the state of i before the increment.
At the same time the actual state of i is incremented. So the value of i is now 1.
bork(0) is invoked.
The exception is thrown.
The exception is handled (by doing nothing)
The current state of i is printed which is equal to 1.

answered Oct 18 '19 at 14:14

Rozart

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That's exactly what I saw in the bytecode as well! I noticed aload, iinc and then the invoke. It's just really counter intuitive. – Jan De Kock Oct 18 '19 at 14:42
https://stackoverflow.com/questions/4445706/post-increment-and-pre-increment-concept - Maybe this topic can give you more insight. Also if you consider my answer helpful, please mark it at as a correct one. :) – Rozart Oct 18 '19 at 14:47

Post increment behavior when throwing an exception

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