Is this an undefined behavior in C?

Question

I am running my C code on gcc to understand pre/post increment operator. However the results that I see are not what I expected. Like for the line 6, since i is 5, it should have been

8 7 6 5 5

But it is 8 7 6 5 8

Then coming to the last line, it displays 14 14 14 14. Can someone please explain this behavior. I had expected 14 14 13 12

Is this compiler dependent? Is the beahviour of printf function on sequence points undefined?

#include <stdio.h>

int main()
{
        i = 5;
        printf("%d %d %d %d %d \n", i, i++, i++, i++, i);
        printf("%d \n", ++i);
        printf("%d \n", ++i);
        printf("%d \n", ++i);
        printf("%d %d %d %d \n", i, ++i, ++i, ++i);

}

It is undefined behaviour. Your expectations are not supported by the C standard — CIsForCookies, Nov 20 '17 at 04:47
It can't be explained, it's undefined behavior. You can't put multiple increments of the same variable into a single statement. — Mark Ransom, Nov 20 '17 at 04:47
Moral of the story: Avoid Pre/post increment in 99% of situations. Readers of your code will appreciate it. — Alexander, Nov 20 '17 at 04:51
@Alexander I appreciate your comment. Intent is to understand how increment operators work, for self evaluation. — Shankhadeep Mukerji, Nov 20 '17 at 04:54
@aschepler: The suggested duplicate is about C++. I'm sure there's a duplicate that's about C, but I haven't found it. — Keith Thompson, Nov 20 '17 at 05:00
@Shankhadeep. Moral of the story is use the operators within their range of applicability and you'll be fine. — Mad Physicist, Nov 20 '17 at 05:02
@ShankhadeepMukerji you will want to carefully read [C11 Standard 6.5.2.4 Postfix increment and decrement operators](http://port70.net/~nsz/c/c11/n1570.html#6.5.2.4) — David C. Rankin, Nov 20 '17 at 05:04

babon · Accepted Answer · 2017-11-20T04:57:20.720

The Standard states that

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be accessed only to determine the value to be stored.

And these are the places where you will find sequence points:

at the end of the evaluation of a full expression (a full expression is an expression statement, or any other expression which is not a subexpression within any larger expression);
at the ||, &&, ?: and comma operators; and
at a function call (after the evaluation of all the arguments, and just before the actual call).

An elaboration of the last point: the comma operators in a function call are not sequence points and the expressions between the ,s can be evaluated in any arbitrary order.

Check this and this for better understanding.

In printf("%d %d %d %d %d \n", i, i++, i++, i++, i);, you are writing to the same memory location more than once between two sequence points, thus invoking undefined behaviour.

@aschepler Thanks for pointing that out. Edited. – babon Nov 20 '17 at 04:54 — babon, Nov 20 '17 at 04:54

Is this an undefined behavior in C?

1 Answers1