Method parameters order in Java

Question

Possible Duplicate:
Order of execution of parameters guarantees in Java?

If I have a Java method like:

    public void func(byte b, byte c) {...}

And I use it like this:

    a = 0;
    func(a++, a);

Wich parameter is passed first? Because if i'm not wrong, if it's the left one then b = 0 and c = 1. And if it's the right one then b = 0 and c = 0?

Thank you.

I think it's an interesting question, but the benefit is near zero. I'd not recommend writing code like this, because it's highly unclear – otherwise you wouldn't have asked. But it shouldn't be that hard to find a solution. — Koraktor, Feb 14 '12 at 23:11
1) I would not want to see this code in production. 2) Test it. Then you'll know. — Anthony Pegram, Feb 14 '12 at 23:11
It's not "passed first", it's "evaluated when". They're passed at the same "time". — Dave Newton, Feb 14 '12 at 23:13
Thank you all for your comments, it was just an existential question :P — 0x77D, Feb 14 '12 at 23:26

score 10 · Accepted Answer · edited Nov 13 '15 at 16:21

10

The arguments are evaluated left to right, as specified in the JLS - section 15.7.4.

edited Nov 13 '15 at 16:21

jchamp

answered Feb 14 '12 at 23:13

MByD

1 Answers1