acetest series

Question

int fun(int n)
{
int table={0,1,1,2,1,2,2,3,1,2,2,3,2,3,3};
int count=0;
for(;n;n>>=4)
count=count+table[n&oxf];
return(count);
}
what will return inn the following function fun()?
a)it returns number of 0's in the number.
b)it returns number of 1's in the number
c)it returns number of 0's and 1's in the number
d)none of the above.

Answer 1

Answer should be option B.

Explanation :

Let n =1000 (8 in decimal)

count = 0

initially in for loop

First iteration

n = 1000

count = count + table [n&oxf]

here anding of n and 15 number is done

1000 & 1111 = 1000

table[8] = 1

hence count = count + 1 => count =1;

Second iteration

initially n is shifted by 4 positions in right = 1000 => 0100 => 0010 => 0001 => 0000 (Here n is 0 hence loop is ended)

and value of count =1 returned.

which is number of 1's in given number (1000 => number of 1's =1)

I've performed this on another number as well it will give number of 1's everytime. :)