gcc compiler printf execution order

Question

For the testing code as follow:

#include <stdio.h>
int addTen(int x, int b[])
{
  b[2] =  x + b[2];
  return b[2];
}
void main(void)
{
   int a[3] = {4,5,6};
   int i = 2;
   printf("%i %i %i \n", a[i], addTen(10,a), a[i]);
}

Why is the output is 16, 16, 6? I know that even if the compiler processes the order from right to left like a[i] <- addTen(10,a) <-a[i]. After calling addTen(10,a), a[i] is already 16 (not 6). So why the output is not 16, 16,16? THANKS!

Presumably because this compiler, at this phase of the moon, evaluates the arguments from right to left as you describe, so `a[i]` is 6 when it evaluates the last one, and 16 when it evaluates the second. But of course this is undefined behaviour, so there isn't an answer. — Mike Seymour, Nov 11 '13 at 17:43

score 4 · Answer 1 · edited May 23 '17 at 12:30

4

It's undefined behavior, you should read about sequence points. You're modifying a and simultaneously reading it in a same expression.

In addition the order of evaluating is not defined.

edited May 23 '17 at 12:30

Community

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answered Nov 11 '13 at 17:40

masoud

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score 4 · Answer 2 · answered Nov 11 '13 at 17:42

4

There's no order defined for evaluating arguments. The compiler is free to evaluate arguments in any order, and will normally choose the most convenient order. So, you can't define any expected output.

answered Nov 11 '13 at 17:42

Filipe Gonçalves

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gcc compiler printf execution order

2 Answers2