Question

How do I count the leading zeroes in an Int32? So what I want to do is write a function which returns 30 if my input is 2, because in binary I have 000...0000000000010.

Answer 1

NOTE Using dotnet core >=3.0? Look here.

Let’s take the number 20 as an example. It can be stated in binary as follows:

    00000000000000000000000000010100

First we "smear" the most significant bit over the lower bit positions by right shifting and bitwise-ORing over itself.

    00000000000000000000000000010100
 or 00000000000000000000000000001010 (right-shifted by 1)
 is 00000000000000000000000000011110

then

    00000000000000000000000000011110
 or 00000000000000000000000000000111 (right-shifted by 2)
 is 00000000000000000000000000011111

Here, because it’s a small number, we’ve already completed the job, but by continuing the process with shifts of 4, 8 and 16 bits, we can ensure that for any 32-bit number, we have set all of the bits from 0 to the MSB of the original number to 1.

Now, if we count the number of 1s in our "smeared" result, we can simply subtract it from 32, and we are left with the number of leading zeros in the original value.

How do we count the number of set bits in an integer? This page has a magical algorithm for doing just that ("a variable-precision SWAR algorithm to perform a tree reduction"… if you get it, you’re cleverer than me!), which translates to C# as follows:

int PopulationCount(int x)
{
    x -= ((x >> 1) & 0x55555555);
    x = (((x >> 2) & 0x33333333) + (x & 0x33333333));
    x = (((x >> 4) + x) & 0x0f0f0f0f);
    x += (x >> 8);
    x += (x >> 16);
    return (x & 0x0000003f);
}

By inlining this method with our "smearing" method above, we can produce a very fast, loop-free and conditional-free method for counting the leading zeros of an integer.

int LeadingZeros(int x)
{
    const int numIntBits = sizeof(int) * 8; //compile time constant
    //do the smearing
    x |= x >> 1; 
    x |= x >> 2;
    x |= x >> 4;
    x |= x >> 8;
    x |= x >> 16;
    //count the ones
    x -= x >> 1 & 0x55555555;
    x = (x >> 2 & 0x33333333) + (x & 0x33333333);
    x = (x >> 4) + x & 0x0f0f0f0f;
    x += x >> 8;
    x += x >> 16;
    return numIntBits - (x & 0x0000003f); //subtract # of 1s from 32
}