Reversing the bits in an integer x

Bit Reversal

I found this code for reversing the bits in an integer x (assume a 32bit value):

unsigned int
reverse(register unsigned int x)
{
x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) <开发者_StackOverflow社区< 8));
return((x >> 16) | (x << 16));
}

I am unable to understand the logic/algorithm behind this code. What is the purpose of all the magic numbers?

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Let's look at how it's done for an 8 bit value:

The first line in the function takes every second bit and moves it left or right:

12345678  -->  1-3-5-7-  -->  -1-3-5-7  -->  21436587
               -2-4-6-8       2-4-6-8-

The second line takes groups of two bits and moves left or right:

21436587  -->  21--65--  -->  --21--65  -->  43218765
               --43--87       43--87--

The third line takes groups of four bits and moves left or right:

43218765  -->  4321----  -->  ----4321  -->  87654321
               ----8765       8765----

Now the bits are reversed. For a 32 bit value you need two more steps that moves bits in groups of 8 and 16.

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That's bit masks. Write them in binary form and remember how bitwise and works. You can also take a piece of paper and write down every step's masks, input and result to sort it out.