[todd8]

Understandably confusing for a non-C-programmer, but it is idiomatic C. It's not code golf; it's just the way one does machine independent bit twiddling.

A mask is used to test each bit position in turn. The tests look like this if written using binary literals:

    0b1000 & x
    0b0100 & x
    0b0010 & x
    0b0001 & x

The '&' is the bitwise AND opperator in C. Of course, we'd have to do as many of these as the word size so 1010111 uses a for loop that starts with the first mask, a 1 in the leftmost position, and shifts it right one position every time through the loop (using the C right shift operator >> on an unsigned mask value). When the one bit is eventually shifted out the right side of the mask, the mask is all zeros so the loop terminates because zero acts like false in the for loop test.

The only other tricky thing is initializing the mask. To set only the leftmost bit in a word the code uses the bit complement operator ~ of C. Breaking it down for a four bit example looks like:

    0u          == 0b0000
    ~0u         == 0b1111
    ~0u >> 1    == 0b0111
    ~(~0u >> 1) == 0b1000

This is the expression that appears in the for loop initializing the mask value, and it works for any word size.

The original article's code was definitely not idiomatic, efficient, or safe (the memory allocation for an array of characters could fail and segfault). The book Hacker's Delight is a great reference for those wanting to understand how to do low level coding, a requirement for close to the hardware work like writing device drivers.