GATE CS 2022 Q11 — Programming in C

Let's trace the execution of the program step-by-step:

1. Initialization:

   int x = 1, z[2] = {10, 11};

   • x is initialized to $1$.
   • Array z is initialized with z[0] = 10 and z[1] = 11.
2. Pointer Assignment and Modification:

   p = &x;
   *p = 10;

   • Pointer p is set to the address of x.
   • Dereferencing p and assigning $10$ updates the value of x to $10$.
3. Pointer Redirection:

   p = &z[1];

   • Pointer p now points to z[1]. Note that p is not used after this statement.
4. Pointer Arithmetic and Modification:

   *(&z[0] + 1) += 3;

   • &z[0] is the address of the first element of the array z.
   • Adding $1$ to this pointer (&z[0] + 1) shifts it to point to the next element, which is &z[1].
   • Dereferencing this expression gives z[1].
   • Thus, the statement is equivalent to: $$z[1] = z[1] + 3 \implies z[1] = 11 + 3 = 14$$
5. Printing the Values:

   printf("%d, %d, %d\n", x, z[0], z[1]);

   • The printed values are: x = 10, z[0] = 10, and z[1] = 14.

Therefore, the output printed is 10, 10, 14, which corresponds to Option D.