#include
int g(int n) { return (n+10); }
int f(int n) { return g(n*2); }
int main() {
int sum, n;
sum=0;
for (n=1; n<3; n++)
sum += g(f(n));
printf("%d", sum);
return 0;
}

The output of the given C program is ______.

The correct answer is 46.

To find the output of the given C program, we can trace its execution step-by-step.

First, let's analyze the functions defined in the program:

1. Function $g\left(n\right)$ takes an integer input $n$ and returns $n+10$.

2. Function $f\left(n\right)$ takes an integer input $n$ and returns $g(n*2)$. Substituting the definition of $g$, we have:

$f\left(n\right)=(n*2)+10=2n+10$

Inside the main function, the variable sum is initialized to 0. A for loop runs for n = 1 and n = 2 (as long as n < 3).

Iteration 1: n = 1

We calculate the value of $f\left(1\right)$:

$f\left(1\right)=g(1*2)=g\left(2\right)=2+10=12$

Now, we compute $g\left(f\right(1\left)\right)=g\left(12\right)$:

$g\left(12\right)=12+10=22$

Updating the sum: sum = 0 + 22 = 22.

Iteration 2: n = 2

We calculate the value of $f\left(2\right)$:

$f\left(2\right)=g(2*2)=g\left(4\right)=4+10=14$

Now, we compute $g\left(f\right(2\left)\right)=g\left(14\right)$:

$g\left(14\right)=14+10=24$

Updating the sum: sum = 22 + 24 = 46.

After the second iteration, the loop terminates. The final value printed is 46.