How are arguments of printf() evaluated?

Question

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int main()
{
    int i = 2;
    printf("%d %d\n", ++i, ++i);
    i = 2;
    printf("%d %d\n", i++, i++);
    return 0;
}

Output:

4 4

3 2

I know that arguments are passed from right to left in printf(). But if I'm getting a 3 2 in the second line, why don't I get a 4 3 in the first line?

The evaluation order of arguments is not right to left (or left to right), it's actually undefined. The compiler may choose to evaluate arguments in whichever order it wants. That goes for *all* function calls. — Some programmer dude, Apr 03 '14 at 11:41
You may also want to read about [Undefined Behavior and Sequence Points](http://stackoverflow.com/q/4176328/440558). — Some programmer dude, Apr 03 '14 at 11:43

score 7 · Answer 1 · edited May 23 '17 at 11:59

7

It is a undefined behavior.

printf("%d %d\n", ++i, ++i);

i is modified more than once in the above statement.

edited May 23 '17 at 11:59

Community

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answered Apr 03 '14 at 11:41

Balu

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Just to clarify my doubts, printf("%d %d\n", i++,i++) will be undefined behavior too na? – user2684198 Apr 03 '14 at 11:46
You might want to read about Sequence Points mentioned above in comments by @Joachim Pileborg – Abhineet Apr 03 '14 at 11:48

score 1 · Answer 2 · answered Apr 03 '14 at 11:58

1

The code you have written comes under Undefined behavior of programming languages.Because the parser of language on each platform could be a different one.Some parser read from left to right and vice-versa.See the wiki link for complete information

answered Apr 03 '14 at 11:58

Ankur

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score 0 · Answer 3 · answered Apr 03 '14 at 13:08

0

You can consider this case with the example of a stack i.e FILO.

In the second printf statement printf("%d %d\n", ++i, ++i);

both ++i will go to the stack one after the other, left to right. since its post-increment, the values that go will be 2 and 2 but while coming out the first value will be 2 and the second value will be 3.

answered Apr 03 '14 at 13:08

Sid

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thus the value printed would be 3 2 – Sid Apr 03 '14 at 13:09

How are arguments of printf() evaluated?

3 Answers3