GATE CS 2004 Q31 — Recursion

To find the value returned by the call f(1), we trace the execution step-by-step. Note that the variable i is declared as static, which means it is initialized only once and retains its updated value across all recursive calls.

Step-by-Step Execution Trace:

1. First Call: f(1)
   • n = 1, and i is initialized to 1.
   • The condition n >= 5 ($1 \ge 5$) is false.
   • n = n + i $\implies n = 1 + 1 = 2$.
   • i++ $\implies i$ becomes 2.
   • Calls f(2).
2. Second Call: f(2)
   • n = 2, and i is currently 2.
   • The condition n >= 5 ($2 \ge 5$) is false.
   • n = n + i $\implies n = 2 + 2 = 4$.
   • i++ $\implies i$ becomes 3.
   • Calls f(4).
3. Third Call: f(4)
   • n = 4, and i is currently 3.
   • The condition n >= 5 ($4 \ge 5$) is false.
   • n = n + i $\implies n = 4 + 3 = 7$.
   • i++ $\implies i$ becomes 4.
   • Calls f(7).
4. Fourth Call: f(7)
   • n = 7.
   • The condition n >= 5 ($7 \ge 5$) is true.
   • The function immediately returns n, which is 7.

Thus, the value returned by f(1) is 7.