How post-increment & pre-increment both are evaluated in function argument?

Question

Possible Duplicate:
post and pre increment in c
Undefined Behavior and Sequence Points

here i want to know why this output comes?? Any one Can Explain me All in proper manner

#include<stdio.h>
   int main() {
   int a=5;
   printf("%d %d %d",a++,a++,++a);
   return 0;
   }

the output of this program is like

In LINUX 7 6 8

http://stackoverflow.com/questions/7606655/pre-fix-and-post-fix-increment-in-c/7606673#7606673 — makes, Oct 24 '11 at 10:48
This http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points actually answers your question. — sharptooth, Oct 24 '11 at 10:48
@Sam_k: means that your code exhibits "undefined behavior" (=as far as the standard is concerned anything can happen), and that this question will be soon closed because there are a million other questions on this same exact topic. — Matteo Italia, Oct 24 '11 at 10:49

score 4 · Accepted Answer · answered Oct 24 '11 at 10:45

4

It's undefined - side effects are only guarantied to be completed at sequence points.

answered Oct 24 '11 at 10:45

Douglas Leeder

score 1 · Answer 2 · answered Oct 24 '11 at 10:45

1

We can't. This is completely compiler dependent in what order the arguments are evaluated.

answered Oct 24 '11 at 10:45

Constantinius

3

Actually behavior is undefined here - `,` in the parameters list is not a sequence point. – sharptooth Oct 24 '11 at 10:47

2 Answers2