C for loop, testing array elements gives weird result

Question

I can't figure out what I did wrong. I want to print non-zero elements and the below code prints none.

#include <stdio.h>

int main ()
{
    int arr[4] = { 0, 3, 0, 7 };
    // print non zero elements
    for (int i = 0; i != 4 && arr[i] != 0; ++i)
        printf("%d\t%d\n", i, arr[i]);
}

But if I move the array test like the below, it works:

#include <stdio.h>

int main ()
{
    int arr[4] = { 0, 3, 0, 7 };
    // print non zero elements
    for (int i = 0; i != 4; ++i) {
        if (arr[i] != 0)
           printf("%d\t%d\n", i, arr[i]);
    }
}

What's wrong?

I think you got it now.

Sumeet
– Sumeet

2015-07-20 02:25:17 +00:00
Commented Jul 20, 2015 at 2:25 — Sumeet
– Sumeet, Commented Jul 20, 2015 at 2:25

Sergey Kalinichenko · Accepted Answer · 2015-07-20 02:10:34Z

6

The reason the first loop is different from the second loop is that the first loop quits as soon as i reaches the position of the first zero in the array, while the second loop always goes all the way up to the forth element.

Your first loop is equivalent to this:

for (int i = 0; i != 4; ++i) {
    if (arr[i] != 0)
       printf("%d\t%d\n", i, arr[i]);
    else
        break;
}

Breaking on seeing zero is what makes the first loop to terminate early.

answered Jul 20, 2015 at 2:10

Sergey Kalinichenko

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2 Comments

BestCoderEver Over a year ago

why would it quit for the first zero in the array? this is not a while loop.

Sergey Kalinichenko Over a year ago

@BestCoderEver The form of the loop doesn't matter: for loop is just like a while loop, but it has two more "compartments" in the loop header - for initialization, and for advancing loop variable. The loop condition in a for loop is the same as the "continuation condition" of a while loop. In your case, that's i != 4 && arr[i] != 0, so your loop would stop at the first zero.

too honest for this site · Accepted Answer · 2015-07-20 02:12:40Z

2

for (int i = 0; i != 4 && arr[i] != 0; ++i)

The loop instantly terminates if arr[i] == 0. Ans as the first element already is 0, it is not even run once.

answered Jul 20, 2015 at 2:12

too honest for this site

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Comments

Sumeet · Accepted Answer · 2015-07-20 02:14:45Z

2

The reason is your first for loop is equivalent to this:

int i=0;
while(i!=4&&arr[i]!=0)
{
  printf("%d\t%d\n", i, arr[i]);
  ++i;
}

That is the condition is being checked in the beginning and therefore your loop terminates before it can begin its first iteration.

answered Jul 20, 2015 at 2:14

Sumeet

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