What is the output of the following C program?#include <stdio.h> int main() { double a[2]={20.0, 25.0}, *p, *q; p = a; q = p + 1; printf("%d,%d", (int)(q - p), (int)(*q - *p)); return 0; }

Let us trace the given C program step-by-step:

1. Array and Pointer Initialization:

   • a is an array of two double elements: a[0] = 20.0 and a[1] = 25.0.

   • p and q are pointers of type double*.

   • p = a; makes pointer p point to the first element a[0] (address &a[0]).

   • q = p + 1; performs pointer arithmetic, making pointer q point to the next double element a[1] (address &a[1]).

2. Evaluating (int)(q - p):

   • In C, pointer subtraction (q - p) between two pointers of the same type yields the number of elements of that base type separating the two pointers, not the byte distance.

   • Since q points to a[1] and p points to a[0], the pointer difference is 1.

   • Casting to int gives 1.

3. **Evaluating (int)(*q - *p):**

   • Dereferencing *q gives 25.0 and *p gives 20.0.

   • The arithmetic difference is $25.0 - 20.0 = 5.0$.

   • Casting to int gives 5.

4. Output:

   • The printf format string is "%d,%d", which prints 1,5.

Hence, the correct option is (B).