What is the output of the following code snippet? #include <stdio.h> #define…

The output of the following C code is: 12 15

Explanation:

short num[3][2] = {3, 6, 9, 12, 15, 18};

This creates a 2D array with 3 rows and 2 columns:

num[0][0] = 3 num[0][1] = 6

num[1][0] = 9 num[1][1] = 12

num[2][0] = 15 num[2][1] = 18

Now evaluate the printf statement:

printf("%d %d", (num+1)[1], *(num+2));

1. *(num+1)[1]

- num points to the first row of the array.

- num + 1 points to the second row → num[1]

- (num+1)[1] refers to num[1][1]

- num[1][1] = 12

- * dereferences the value, so the result is 12

2. **(num+2)

- num + 2 points to the third row → num[2]

- *(num+2) gives the base address of row num[2]

- **(num+2) accesses the first element of that row

- num[2][0] = 15

Hence, the final output is:

12 15