ISRO2020-59

Question

Consider the following recursive C function that takes two arguments

unsigned int rer(unsigned int n, unsigned int r){
    if(n>0)return(n%r + rer(n/r,r));
    else retturn 0;
}

What is the return value of the function $rer$ when it is called as $rer(513,2)$?

• A. $9$
• B. $8$
• C. $5$
• D. $2$

Comments

The correct answer  is $(d).$

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The code basically counts the number of 1's in the binary representation of 513.

Answer 1

Option d) is correct

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$rer(513,2)=1+rer(256,2)=2$

$rer(256,2)=0+rer(128,2)=1$

$rer(128,2)=0+rer(64,2)=1$

$rer(64,2)=0+rer(32,2)=1$

$rer(32,2)=0+rer(16,2)=1$

$rer(16,2)=0+rer(8,2)=1$

$rer(8,2)=0+rer(4,2)=1$

$rer(4,2)=0+rer(2,2)=1$

$rer(2,2)=0+rer(1,2)=1$

$rer(1,2)=1+rer(0,2)= 1$

$rer(0,2)=return \ 0$

Answer 2

If we observe it carefully,the code counts the no of 1's in the binary representation of the no 513,so we can write its binary representation and see how many 1's are there. It will come out to be 2

Answer 3

what you can observe from the above code is that n%2 will extract the last digit of n and n/2 will extract the first

(no. of digits in the binary representation of the number)-1

Binary representation of the number is 1000000001. It has 2 1s so addition of 1s will take place 2 times

so d)2