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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