GATE CS 2016 Set 1 Q12 — Programming in C

The function is declared as:

void f(int, short);

This means f accepts two arguments: the first of type int, the second of type short, and it returns void (no value).

Let's analyze the variables declared in main():

• int i = 100; — i is of type int.
• short s = 12; — s is of type short.
• short *p = &s; — p is a pointer to short, so *p is of type short.

Now, let's evaluate each option:

• Option A: f(s, *s)
  Here, s is a short variable, not a pointer. Applying the dereference operator * to a non-pointer variable (*s) is a compilation/type checking error.
• Option B: i = f(i, s)
  The function f has a return type of void. Attempting to assign a void return value to an int variable i results in a type checking error.
• Option C: f(i, *s)
  Similar to Option A, *s is invalid because s is not a pointer.
• Option D: f(i, *p)
  
  • The first argument i is of type int, which matches the first parameter.
  • The second argument *p dereferences the pointer p (of type short*), yielding a value of type short, which matches the second parameter.
  • The function call is not assigned to any variable, which is correct since f returns void.
  This expression is completely valid and will not result in any type checking error.