Yet an other i = i++

Question

I have encountered a sorcery today.

Language: C

Code:

#include <stdio.h>

main()
{
    int i = 5;
    i = i++;
    printf ("%i", i);
}

Output:

How? Why?

This is supposed to be tricky code but the other way around. The negligent programmer would think that i = i++ is just simple increment but it's not. Yet it works like one here. It's supposed to be 5! Like in JavaScript.

What should be happening.

i gets the value of 5.
i++ returns 5.
i is post incremented by i++ (to 6).
i gets the value of 5 (returned by i++).
the value of i (5) gets printed.

Yet it is 6.

I haven't been able to find a description to this on SO, or the whole internet (just the other way around).

What is broken here?

Please explain.

This is an undefined behaviour. See here for reference - http://stackoverflow.com/questions/949433/could-anyone-explain-these-undefined-behaviors-i-i-i-i-i-etc — Sukrit Kalra, May 12 '13 at 20:01
@SukritKalra Thanks for the link, the answers there are quite descriptive, just what I was looking for. — vinczemarton, May 12 '13 at 20:12

score 11 · Accepted Answer · answered May 12 '13 at 20:01

11

It is undefined behavior to store more than once to an object without an intervening sequence point.

In particular, your step 3 and 4 have no defined ordering, the increment (and store) or the store could happen first.

answered May 12 '13 at 20:01

luser droog

18,988
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I'll accept this after reading the wikipedia article about sequence points. – vinczemarton May 12 '13 at 20:16

Yet an other i = i++

1 Answers1