8,690 views

Consider the function func shown below:

int func(int num) {
int count = 0;
while (num) {
count++;
num>>= 1;
}
return (count);
} 

The value returned by func($435$) is ________

will left shift will in multiplication?

yes
Operator is right shift. >>

Answer is $9$.

$435-(110110011)$

num $>>=$ $1$; implies a num is shifted one bit right in every while loop execution.While loop is executed $9$ times successfully and $10$th time num is zero.

So count is incremented $9$ times.

Note:

Shifting a number "1"bit position to the right will have the effect of dividing by $2$:

8 >> 1 = \$4    // In binary: (00001000) >> 1 = (00000100)
by

edited by
doubt: why we are not considering the storing technique like little endian or big endian.. in C int take 16 bits if we consider big endian [uper bytes take lower address and lower bytes take upper address] then answer is fine but what if little endian...?
We can count the number of set bits(1) by doing simple modifications in the code.

If(num &1) set_bits++; // By checking LSB 1 or not.

else reset_bits++;

total bits=set_bits+reset_bits;
int func(int num) // take decimal number
{
int count = 0;
while (num) // until all bits are zero
{
count++; // count bit
num>>= 1; // shift bits, removing lower bit
}
return (count); // returns total number of bits
} 

(435)10 = (110110011)2
So, the given program counts total number of bits in binary representation . Hence, answer is 9

http://stackoverflow.com/questions/109023/how-to-count-the-number-of-set-bits-in-a-32-bit-integer

The function mainly returns position of Most significant bit in binary representation of n. The MSB in binary representation of 435 is 9th bit.

Another explanation : >> in right shift. In other words, it means divide by 2. If keep on dividing by 2, we get: 435, 217, 108, 54, 27, 13, 6, 3, 1. Therefore, the count is 9.

### 1 comment

MSB not MSD

435 in binary ....................100110011...............count =0

first count is incremented then bitwise shift >>1 is done

count = 1 ............010011001

count = 2 ............001001100

count = 3 ............000100110

count = 4 ............000010011

count = 5 ............000001001

count = 6 ............000000100

count = 7 ............000000010

count = 8 ............000000001

count = 9 ............000000000

finally, count = 9