GATE | GATE CS 2020 | Question 32

• Difficulty Level : Medium
• Last Updated : 11 Aug, 2021

Consider the following C program.

 `#include  ``int` `main ()  { ``    ``int`  `a[4][5] = {{1, 2, 3, 4, 5}, ``                    ``{6, 7, 8, 9, 10}, ``                    ``{11, 12, 13, 14, 15}, ``                    ``{16, 17, 18, 19, 20}}; ``    ``printf``(``"%d\n"``, *(*(a+**a+2)+3)); ``    ``return``(0); ``}  `

The output of the program is _______ .

Note – This question was Numerical Type.
(A) 18
(B) 19
(C) 20
(D) 14

Explanation: Given a[4][5] is 2D array. Let starting address (or base address) of given array is 1000.

Therefore,

```= *(*(a+**a+2)+3))
= *(*(1000+**1000+2)+3))
= *(*(1000+3)+3))   {given element **1002 = 3}
= *(*(1003)+3))
= *((1003)+3)    {4th row selected in given matrix}
= *((1003)+3)     {address of 4th element in 4th row}
= a[3][3]
= 19            {element selected a[3][3] = 19} ```
 `#include  ``int` `main ()  { ``    ``int`  `a[4][5] = {{1, 2, 3, 4, 5}, ``                    ``{6, 7, 8, 9, 10}, ``                    ``{11, 12, 13, 14, 15}, ``                    ``{16, 17, 18, 19, 20}}; ``    ``printf``(``"%d\n"``, *(*(a+**a+2)+3)); ``    ``return``(0); ``}  `

Option (B) is correct.

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