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

Anyone know how to see the output of what the calc() function is returning??

I know it will be very huge number but even if I change to long long int then also not able to see output. just wanted to know is there is any facality in C to see that much huge no. below is my code feel free to edit and let me know if you are able to see output. cause it is outputing 0 and i think this will going off the double doube int max limit.

https://ideone.com/H3FCkD
 

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Just count the number of digits in the result. And a double value can hold how many digits?

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thanks this works...🙂

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?

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);
}

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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.