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

Question

Why does this code shows the output as "3 2" instead of "2 3" ?

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<string>
#include<map>
#include<vector>
using namespace std;
int main()
{

    int i=2;// declare

    printf("%d %d\n",i++,i++);//print
    return 0;
}

Output is : "3 2" Why it prints in reverse order

Take a look at this http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points — suspectus, Jan 19 '14 at 18:06
Can I ask what prompted you to ask this question? If it wasn't a discussion about its strange behaviour, what was it? — Joseph Mansfield, Jan 19 '14 at 18:09
@JosephMansfield: There's a daily "who will ask this question today?" lottery, and it was this guy's turn to be picked. — Lightness Races in Orbit, Jan 19 '14 at 18:10
SO should add these questions to the list [_"What types of questions should I avoid asking ?"_](http://stackoverflow.com/help/dont-ask) — P0W, Jan 19 '14 at 18:15

score 4 · Answer 1 · answered Jan 19 '14 at 18:13

In this statement, the expression "printf(...)" modifies the variable "i" more than once without an intervening sequence point.

This behavior is undefined.

The compiler has detect a case where the same variable has been modified more than once in an expression without a sequence point between the modifications. Because what modification will occur last is not defined, this expression might produce different results on different platforms.

Rewrite the expression so that each variable is modified only once.

even you might get output "2 3" in different compiler

Rakesh KR · Answer 2 · 2014-01-19T18:21:03.857

The order of evaluation of printf is from right to left in here

First evaluate

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

Then

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

So you got the output as 3 2

The behaviour will be definitely undefined due to the undefined evaluation order of parameters.

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 read only to determine the value to be stored.

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

2 Answers2

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