What happens if you try to compile and run this program?

In the f1 function, n = n * n; squares the input n and assigns the result back to n.

In the f2 function, f1(n) * f1(n) calls f1 twice with the input n and multiplies the results.

In the main function, printf("%d \n", f2(1)); prints the result of f2(1).

Let's calculate:

1.f1(1) returns 1 * 1 = 1.

2.f2(1) calls f1 twice with the input 1, so it's f1(1) * f1(1) = 1 * 1 * 1 * 1 = 1.