How do I count the number of zero bits in an integer?

Question

How would i go about finding the number of \'zero\' bits in C++. Suppose I have an integer;

int value = 276; 

For which I have the bits 100010

Answer 1

If you want efficiency then there is a good implementation in the book "Hackers Delight"

22 instructions branch free.

unsigned int count_1bits(unsigned int x)
{
    x = x - ((x >> 1) & 0x55555555);
    x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
    x = x + (x >> 8);
    x = x + (x >> 16);
    return x & 0x0000003F;
}

unsigned int count_0bits(unsigned int x)
{
    return 32 - count_1bits(x);
}

I'll try to explain how it works. It is a divide-and-conquer algorithm.

(x >> 1) & 0x55555555

Shifts all bits 1 step to the right and takes the least significant bit of every bit pair.

0x55555555 -> 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 01 (16x2 bit pairs)

So basically you will have the following table of all 2 bit permutations.

1. (00 >> 1) & 01 = 00
2. (01 >> 1) & 01 = 00
3. (10 >> 1) & 01 = 01
4. (11 >> 1) & 01 = 01

x - ((x >> 1) & 0x55555555);

Then you subtract these from the non shifted pairs.

1. 00 - 00 = 00 => 0 x 1 bits
2. 01 - 00 = 01 => 1 x 1 bits
3. 10 - 01 = 01 => 1 x 1 bits
4. 11 - 01 = 10 => 2 x 1 bits

x = x - ((x >> 1) & 0x55555555);

So now we have changed every 2 bit pair so that their value is now the number of bits of their corresponding original 2 bit pairs... and then we continue in similar way with 4 bit groups, 8 bit groups, 16 bit groups and final 32 bit.

If you want a better explanation buy the book, there are a lot of good explanation and discussions of alternative algorithms etc...