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Consider the following code (which does not do anything significant)int h(int p) {    return p++;}int g(int x, int y) {    int z;   return h(y);}int f(int a, int b) {    if(a % 3 == 0) {        g(a, 2);        } else {        h(a);    }       return 10;}int main() {    int m;    scanf("%d", &m);    if(m % 2 == 0) {        f(m, 20);    } else {        g(20, m);    }       return 0;   }How many functions can be active the same time, in the execution of the above program?

Question

Consider the following code (which does not do anything significant)int h(int p) {    return p++;}int g(int x, int y) {    int z;   return h(y);}int f(int a, int b) {    if(a % 3 == 0) {        g(a, 2);        } else {        h(a);    }       return 10;}int main() {    int m;    scanf("%d", &m);    if(m % 2 == 0) {        f(m, 20);    } else {        g(20, m);    }       return 0;   }How many functions can be active the same time, in the execution of the above program?

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Solution 1

The maximum number of functions that can be active at the same time in the execution of the above program is 3. Here's the step-by-step explanation:

  1. The main() function is always active as it is the entry point of the program.

  2. If the input m is even, the main() function calls the f() function. Inside f(), if a (which is m in this case) is divisible by 3, it calls the g() function.

  3. Inside the g() function, it calls the h() function.

So, in this scenario

This problem has been solved

Solution 2

The maximum number of functions that can be active at the same time in the execution of the above program is 3. This happens when the main function calls the function f (when m is even), and then function f calls function g (when a is not divisible by 3), and then function g calls function h. So, the functions main, f, g, and h can all be active at the same time. However, since each function returns before the next one is called, only one function is actually executing at any given time.

This problem has been solved

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