GATE CSE 2018 | Question: 21

Question

Consider the following $\text{C}$ program:

#include<stdio.h>

int counter=0;

int calc (int a, int b) {
        int c;
        counter++;
        if(b==3) return (a*a*a);
        else {
                c = calc(a, b/3);
                return (c*c*c);
        }
}

int main() {
        calc(4, 81);
        printf("%d", counter);
}

The output of this program is ______.

Comments

Instead of global if count definition is inside calc then answer would be an error? Please anyone if explain.

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@Tushar Rana if counter variable is inside calc then - 

1. It would reset to zero inside every iteration of calc()
2. We cannot call it inside main() as it would become a local variable

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Note to self and all fellow aspirants:

These kinds of questions only test our presence of mind while reading the question.

If we carefully see that calc() is never printing anything, and the only thing that matters is the value of the global variable counter,

So, we can skip all the hassle of calculating everything and how many times counter++ is executed

 

Answer 1

int main() {
        calc(4, 81);
        printf("%d", counter);
}

printf("%d", counter);
So we need only counter value.

Each function increments counter value by 1. Goal is to find the number of function calls.
Squence of function calls:
calc(4, 81) ---> calc(4, 27) ---> calc(4, 9) ---> calc(4, 3) ---> return

4 function calls.

counter = 4

Comments

For given question calc(4,81) will return 4^81 to main().

 

And if in question return(c+c+c) then calc(4,81) will return 1728 to main(). And variable c local or global doesn’t matter(same output 1728). @Abhineet Singh

https://ideone.com/dBkD3P

 

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Yes it is returning 4^81

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@HHH777

by default function is integer only so no error.

Answer 2

There is no need to calculate return value we just need value of counter.

So answer is 4.

Comments

why it's return value is displaying 0? shouldn't it be 64?

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we need not return c, we need to return the value of the counter

 

Answer 3

#include<stdio.h>

int counter=0;

int calc (int a, int b) {
        int c;
        counter++;
        if(b==3) return (a*a*a);
        else {
                c = calc(a, b/3);
                return (c*c*c);
        }
}

int main() {
        calc(4, 81);
        printf("%d", counter);
}

---

Here counter is a global variable, you can use anywhere of the program, 

 Now when main() call the calc() by the value (4,81) 

--> counter=counter+1 so now counter=1.

Now for 81 not eq to 3 so else block got executed and Now b=81/3=27.

Again counter increased by 1 and now counter =2.

In this way when b=3 counter will become 4 and cacl() return 64;

counter=3

int calc(4,3)

{

int c;

counter++;

if(b==3) return (a*a*a);
        else {
                c = calc(a, b/3);
                return (c*c*c);
        }

}

So, ans is 4.

Answer 4

The main function is just printing counter value. There's no need to calculate the values of return statements in the function calc().

 

Therefore , Answer is 4.

Answer 5

we just need to count the no,, of function calls have been made to determine the value of counter

i.e calc(4,81) - function call no 1

     calc(4,27) - function call no 2

     calc(4,9) - function call no 3

     calc(4,3) - function call no 4

since each function call increments the value of counter variable by 1 and it is a global varible which is stored in static memory section it will retain it's value.

Therefore, the value of counter =4

Answer 6

 

4 function calls.

counter = 4