Counting the number of ones in the binary representation of an int.

Here are the 32 bits of a normal int

    n = ABCDEFGHIJKLMNOPQRSTUVWXYZabcdef
    
0x55555555 = 01010101010101010101010101010101

n & 0x55555555  and  (n >> 1) & 0x55555555  are

     0B0D0F0H0J0L0N0P0R0T0V0X0Z0b0d0f
and  0A0C0E0G0I0K0M0O0Q0S0U0W0Y0a0c0e

Add those two together, and all the bit sums A+B,
C+D, E+F, G+H, ..., e+f will be 0, 1, or 2, and will
fit exactly into the two bits available to them

So after
   n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
   
n could be written as
     AACCEEGGIIKKMMOOQQSSUUWWYYaaccee
where
   AA represents the original bits A+B
   CC represents the original bits C+D
   EE represents the original bits E+F
etc.

Next

0x33333333 = 00110011001100110011001100110011

n & 0x33333333  and  (n >> 2) & 0x33333333  are

     00CC00GG00KK00OO00SS00WW00aa00ee
and  00AA00EE00II00MM00QQ00UU00YY00cc
 
Add those two together, and all the bit sums AA+CC,
EE+GG, II+KK, ..., ee+cc will be 0, 1, 2, 3, or 4 and 
will fit easily into the four bits available to them

So after
   n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
   
n could be written as
     AAAAEEEEIIIIMMMMQQQQUUUUYYYYcdef
where
   AAAA represents the original bits A+B+C+D
   EEEE represents the original bits E+F+G+H
   IIII represents the original bits I+J+K+L
etc.
 
just repeat three more times for a total of

   n = (n & 0x55555555) + ((n >> 1) & 0x55555555);
   n = (n & 0x33333333) + ((n >> 2) & 0x33333333);
   n = (n & 0xFFFFFFFF) + ((n >> 4) & 0xFFFFFFFF);
   n = (n & 0x00FF00FF) + ((n >> 8) & 0x00FF00FF);
   n = (n & 0x0000FFFF) + ((n >> 16) & 0x0000FFFF);

and n ends up as the sum of its own original bits.