GATE CS 2013 Q42 — Programming in C

To find the return value of $f(p, p)$ with $p$ initialized to $5$, we analyze the parameter passing mechanisms:

• The first parameter x is passed by reference, meaning x is an alias for the variable p. Any modification to x directly updates p.
• The second parameter c is passed by value, meaning c is a local copy of p's value at the time of the call.

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

1. Call 1: $f(p, 5)$ where $p = 5$
   • c becomes $4$.
   • x = x + 1 increments $p$ to $6$.
   • Returns f(x, 4) * x.
2. Call 2: $f(p, 4)$ where $p = 6$
   • c becomes $3$.
   • x = x + 1 increments $p$ to $7$.
   • Returns f(x, 3) * x.
3. Call 3: $f(p, 3)$ where $p = 7$
   • c becomes $2$.
   • x = x + 1 increments $p$ to $8$.
   • Returns f(x, 2) * x.
4. Call 4: $f(p, 2)$ where $p = 8$
   • c becomes $1$.
   • x = x + 1 increments $p$ to $9$.
   • Returns f(x, 1) * x.
5. Call 5: $f(p, 1)$ where $p = 9$
   • c becomes $0$.
   • Since c == 0, it returns $1$.

Now, we unwind the recursion. Since x is a reference to p, and the value of p is $9$ at the end of the recursion, any evaluation of x during the multiplication phase will use the current value of $p$, which is $9$:

• Call 4 returns: $1 \times 9 = 9$
• Call 3 returns: $9 \times 9 = 81$
• Call 2 returns: $81 \times 9 = 729$
• Call 1 returns: $729 \times 9 = 6561$

Thus, the final return value is 6561.