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What will be the output of the program ?

#include <stdio.h>

int main()
{
    static int arr[] = {0, 1, 2, 3, 4};
    int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};
    int **ptr=p;
    ptr++;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    *ptr++;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    *++ptr;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    ++*ptr;
    printf("%d, %d, %d\n", ptr-p, *ptr-arr, **ptr);
    return 0;
}

 

  • A.0, 0, 0
    1, 1, 1
    2, 2, 2
    3, 3, 3
  • B.1, 1, 2
    2, 2, 3
    3, 3, 4
    4, 4, 1
  • C.1, 1, 1
    2, 2, 2
    3, 3, 3
    3, 4, 4
  • D.0, 1, 2
    1, 2, 3
    2, 3, 4
    3, 4, 5

Answer: C

Correct answer: 1, 1, 1
2, 2, 2
3, 3, 3
3, 4, 4

p holds the five addresses arr … arr+4, and ptr walks along p. Take the four statements in turn.

  • ptr++ - ptr = p+1. Then ptr-p is 1, *ptr-arr is 1, **ptr is arr[1] = 1 → 1, 1, 1
  • *ptr++ - the ++ binds to ptr, and the dereference is thrown away, so ptr = p+2 → 2, 2, 2
  • *++ptr - again only ptr moves, to p+3 → 3, 3, 3
  • ++*ptr - here the increment applies to *ptr, that is to p[3], which becomes arr+4. ptr itself has not moved → ptr-p = 3, *ptr-arr = 4, **ptr = arr[4] = 4 → 3, 4, 4

The point of the question is that the last line increments what the pointer points to, while the three before it increment the pointer.

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