post increment and pointer in printf [duplicate]

Question

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Why are these constructs using pre and post-increment undefined behavior? (14 answers)

Closed 8 years ago.

Can someone please explain why this code is outputing 2 1, I thought post-increment was supposed to be applied after the printf instruction.

#include <stdio.h>

int main() {
  int i=1;
  int *p=&i;
  printf("%d %d\n", *p ,i++);

  return 0;
}

Answer 1

The order of evaluation of is not specified for this case in the standard and so you can not determine whether *p or i++ will be evaluated first. The C99 draft standard says in section 6.5.2.2 Function calls paragraph 10 says:

The order of evaluation of the function designator, the actual arguments, and subexpressions within the actual arguments is unspecified, but there is a sequence point before the actual call.

This is also undefined behavior because you are modifying i and accessing the previous value of i in another expression within the same sequence point, the draft standard in section 6.5 Expressions paragraph 2 says:

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression.72) Furthermore, the prior value shall be read only to determine the value to be stored

Answer 2

Comma is a sequence point, but not in function calls.
And printf is a function with arguments separated with commas.
In expression like this:
if(i+2,i++). . . comma is a sequence point and assures that expressions will be evaluated from left to right, and final value of expression will be value of rightmost sub-expression. In this case i++.
On the other hand in funcion call:
function(i+1,i++,i--) comma in not a sequence point and any of sub-expressions can be evaluated as first. The only thing sure is that they all be evaluated, but not in any specific order.