Question 35

We are given two mutually recursive functions:

void fun1(int n) {
    if (n == 0) return;
    printf("%d", n);
    fun2(n - 2);
    printf("%d", n);
}

void fun2(int n) {
    if (n == 0) return;
    printf("%d", n);
    fun1(++n);
    printf("%d", n);
}

Trace the execution starting from fun1(5):

1. fun1(5): prints 5, calls fun2(3).
2. fun2(3): prints 3, increments $n$ to 4 (++n), calls fun1(4).
3. fun1(4): prints 4, calls fun2(2).
4. fun2(2): prints 2, increments $n$ to 3 (++n), calls fun1(3).
5. fun1(3): prints 3, calls fun2(1).
6. fun2(1): prints 1, increments $n$ to 2 (++n), calls fun1(2).
7. fun1(2): prints 2, calls fun2(0).
8. fun2(0): $n == 0$, returns immediately without printing.

Now trace the returns from the call stack:

• fun1(2) resumes: prints its local $n = 2$, returns.
• fun2(1) resumes: its local $n$ was incremented to 2, so prints 2, returns.
• fun1(3) resumes: prints its local $n = 3$, returns.
• fun2(2) resumes: its local $n$ was incremented to 3, so prints 3, returns.
• fun1(4) resumes: prints its local $n = 4$, returns.
• fun2(3) resumes: its local $n$ was incremented to 4, so prints 4, returns.
• fun1(5) resumes: prints its local $n = 5$, returns.

Concatenating the printed digits in order gives:
5 3 4 2 3 1 2 2 2 3 3 4 4 5
which is 53423122233445 (Option A).