[FIXED] why is output 2 3 0?

Issue

I'm solving some exam questions, and I came across this one and I just don't get it.

   int main()
    {
       int a[3+2] = {1, 2, 3}, i;
       for(i = 0; a[i++]++;)
          printf("%d", a[i]);
    return 0;
    }

Why is output of this code 2 3 0? I tried few ways of thinking but cant undertsand it.

Solution

In first iteration, a[i++]++ will do two things:

1. Return value of i and increment it (first ++), thus leaving us with a[0] (i initialised as 0) as of now,
2. Now coming to the a[0], it's post-incremented again (second ++), so this finally returns a[0] and then increments it.

So for the first iteration, the value returned in for's condition is a[0] which is 1, thus the loop will proceed to its body.

Coming to the body of our for loop, we now have two changes from the above statement, i and a[i] both incremented, so we have i = 1 and a[0] = 2 (a[0] was incremented by 1, remember the outer ++).

Continuing in the same fashion, we will move on to i = 2, a[1] = 3 (a[1] was 2 but post-incremented, so became 3).

After this, again coming back to the loop's conditional, we have i (=2) and a[i] (=3) both post-incremented again, so the value returned will be a[i] that is a[2] (since i is 2 as of now, now it is incremented after being returned), so i would be 3 in the next statement inside for loop's body, which would make the array element as a[3] that is 0 (that's why you get 0 in the last iteration).

Since i = 3 and from the array declaration, we have the default value of a[3] as 0, it gives i = 3, a[i] = a[3] = 0 in the next iteration.

Again coming back to the loop conditional, it will return a[i] which at this point is a[3] that is 0, thus failing the condition and hence halting the loop. Hence, you don't print anything afterwards.

Answered By - Jarvis

Answer Checked By - Mary Flores (PHPFixing Volunteer)