Output of C Program | Set 26

Predict the output of following C programs.

Question 1

#include <stdio.h>

int main()
{
  int arr[] = {};
  printf("%d", sizeof(arr));
  return 0;
}

Output: 0
C (or C++) allows arrays of size 0. When an array is declared with empty initialization list, size of the array becomes 0.

Question 2

#include<stdio.h>

int main()
{
  int i, j;
  int arr[4][4] = { {1, 2, 3, 4},
                    {5, 6, 7, 8},
                    {9, 10, 11, 12},
                    {13, 14, 15, 16} };
  for(i = 0; i < 4; i++)
    for(j = 0; j < 4; j++)
      printf("%d ", j[i[arr]] );

  printf("\n");

  for(i = 0; i < 4; i++)
    for(j = 0; j < 4; j++)
      printf("%d ", i[j[arr]] );

  return 0;
}

Output:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 5 9 13 2 6 10 14 3 7 11 15 4 8 12 16

Array elements are accessed using pointer arithmetic. So the meaning of arr[i][j] and j[i[arr]] is same. They both mean (arr + 4*i + j). Similarly, the meaning of arr[j][i] and i[j[arr]] is same.

Question 3

#include<stdio.h>
int main()
{
    int a[2][3] = {2,1,3,2,3,4};
    printf("Using pointer notations:\n");
    printf("%d %d %d\n", *(*(a+0)+0), *(*(a+0)+1), *(*(a+0)+2));
    printf("Using mixed notations:\n");
    printf("%d %d %d\n", *(a[1]+0), *(a[1]+1), *(a[1]+2));
    return 0;
}

Output:

Using pointer notations:
2 1 3
Using mixed notations:
2 3 4 

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