Question

#include<stdio.h>

#define N (sizeof(array) / sizeof(array[0]))

main()
{
   int array[5]={1,2,3,4,5};
   int d;
   for(d=-1;d <= N;d++)
       printf("%d\n",array[d+1]);
   return 0;
}

The above code is not displaying anything? Can anybody tell me why?

Answer 1

First, I’ll note in passing that the code as it stands will read outside the bounds of the array, since d+1 will be 6 at the last iteration of the loop, while the highest valid array index is 4. But this isn’t the main problem with your code; as it stands, when d = -1, the condition d <= N will actually evaluate to false, and thus the loop terminates at once without going through any iterations. The problem is that the result of the expression sizeof(array) / sizeof(array[0]) is of type size_t, which is an unsigned integer, while d is an int, i.e. a signed integer. Prior to the actual comparison, d is converted to a size_t (this is called type promotion), resulting in a large positive integer, and thus the expression d <= N evaluates to false. See this c-faq entry for more information. To really drive the point home, you may want to try the following code, and see if it works as expected:

int d = -1;
printf("%u\n", d); 

size_t n = 5000;
if (d > n) printf("oops!\n");

Fixing your code is fairly simple – for example, as others have suggested, rewriting the loop as

for (d = 0; d < N; d++) {
        printf("%d\n", array[d]);    
}

will work.