# Output of C Program | Set 26

• Difficulty Level : Medium
• Last Updated : 27 Dec, 2016

Predict the output of following C programs.

Question 1

 `#include ``  ` `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`` ` `int` `main()``{``  ``int` `i, j;``  ``int` `arr = { {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``int` `main()``{``    ``int` `a = {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+0), *(a+1), *(a+2));``    ``return` `0;``}`

Output:

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

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