GATE IT 2005 | Question: 57

Question

What is the output printed by the following program?

#include <stdio.h>

int f(int n, int k) {
    if (n == 0) return 0;
    else if (n % 2) return f(n/2, 2*k) + k;
    else return f(n/2, 2*k) - k;
}

int main () {
    printf("%d", f(20, 1));
    return 0;
}

• A. $5$
• B. $8$
• C. $9$
• D. $20$

Comments

Answer should be -15 which is not given in the options, observe carefully the value of k

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int f(int n, int k) {
    if (n == 0) return 0;
    else if (n % 2) return f(n/2, 2*k) + k; // This condition is true when n%2 is 1 
    else return f(n/2, 2*k) - k; // otherwise
}

bthebestSelf you are doing n%2 = 0 as true, then you get -15 as the answer. If you doing n%2 = 0 as false then you get 9 as an answer :)

Answer 1

See the following calling sequence. Boxed values show the return values.

Hence, answer is option C.

Comments

ya, its ok. you mistook, you will never mistook it agin.

ok keep it up

vamsi

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@Vamsi Krishna Satya  I also mistook...

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after much struggle finally understood

Answer 2

The sequence has to be followed.

6.) f(20,1) = 9.

5.) f(10,2) - 1 = 9

4.) f(5,4) - 2 = 10

3.) f(2,8) + 4 = 12

2.) f(1,16) - 8 = 8

1.) f(0,32) + 16 = 16

Comments

Why have u executed else part in 5th step I guess else if has to be executed since 10 which is divisible by 2.

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else if (n%2)

This will be true if n is NOT a multiple of 2. When n is a multiple of 2, n%2 returns 0 and if will be false. 

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Thank you sir.

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how values are come i mean like 9,10,8.,16 etc.Anybody plz explain?

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Can anybody plzz explain how this series values comes.

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its clear now.

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Else if (n%2) when n is multiple of 2 it will be like else if (0) therefore it will return false and will execute next else and   return f(n/2,2*k) - K....

Answer 3

9 will be printed.

Answer: C