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Question


What will be the output of the program if the array begins at 65472 and each
integer occupies 2 bytes?


#include<stdio.h>
int main()
{
int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
printf("%u, %u\n", a+1, &a+1);
return 0;
}
Options:
A .  65474, 65476
B .  65480, 65496
C .  65480, 65488
D .  65474, 65488
Answer: Option B

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer

 array having the 3 rows and 4 colums dimensions.

Step 2: printf("%u, %u`setminus`n", a+1, &a+1);

The base address(also the address of the first element) of array is 65472.

For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore,

 a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 +

 (4 ints * 2 bytes) = 65480

Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes

 "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496

Hence the output of the program is 65480, 65496




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