What is the return value of f(p, p), if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value.

int f (int &x, int c) {
c = c - 1;
if (c == 0) return 1;
x = x + 1;
return f(x, c) * x;
}

Question

What is the return value of f(p, p), if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value.

int f (int &x, int c) {
 c = c - 1;
 if (c == 0) return 1;
 x = x + 1;
 return f(x, c) * x;
}

Accepted Answer

6561. x is passed by reference to p throughout, while c (by value) counts 5->4->3->2->1->0, incrementing the shared p four times from 5 to 9. The base case returns 1, and each of the four returns multiplies by the final shared value 9, giving 9^4 = 6561.

What is the return value of f(p, p), if the value of p is initialized to 5 before the call? Note that the first parameter is passed by reference, whereas the second parameter is passed by value. int f (int &x, int c) { c = c - 1; if (c == 0) return 1; x = x + 1; return f(x, c) * x; }

Solution

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