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Question

void foo(int n)

{

  while (n! = 0)

       {

         if  (!(n & 1))

          printf(“*”);

          n = n >> 1 ;

     }

}

 

The number of times printf(“*”) is executed, when the value 2 raised to power 24 is passed to the function foo() is ?

Comments

if condition satisfies if the LSB=0, after that n right shift by 1 bit

So, in 2$^{24}$ ==> 100000..... ===> 1 followed by 24 zero's

∴Total 24 times * printed.

 

If you didn't understood this, then take a small example 8 or 16, and analyse the code line by line, then read my comment again, then you will understood it clearly!

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Thanks i got it

Answer 1

24 is the answer

Last statement is right shift operator which means divide by 2 and the condition is bitwise AND with 1 which keep on resulting zero (!0=1 condition true) till the number is greater than 1. At 1 (1&1=1 condition becomes false) you can run the code for small example like 2^3 and printf will be executed 3 times.

Comments

Thanks i got it

Answer 2

Ans is 24.

First understand following operation on bits

n & 1 returns true when last bit of n is 1 or say when s is odd. Therefore !(n&1) returns true when n is even or last bit of n is 0.

n= n >>1, shifts the bits of n rightwards by 1 position. It is equivalent to n = n/2

Now $2^{24}$ means 1 followed by 24 zeros in binary.

As loop executes n will keep shifting rightwards..

$2^{24}$ -> $2^{23}$ -> $2^{22}$ ->... -> $2^{1}$

Therefore 24 times.

Comments

Thanks for the detailed explanation