How do the bit manipulations in this bit-sorting code work?

Question

Jon Bentley in Column 1 of his book programming pearls introduces a technique for sorting a sequence of non-zero positive integers using bit vectors.

I have taken the p

Answer 1

The first 3 constants are inter-related. BITSPERWORD is 32. This you'd want to set based on your compiler+architecture. SHIFT is 5, because 2^5 = 32. Finally, MASK is 0x1F which is 11111 in binary (ie: the bottom 5 bits are all set). Equivalently, MASK = BITSPERWORD - 1.

The bitset is conceptually just an array of bits. This implementation actually uses an array of ints, and assumes 32 bits per int. So whenever we want to set, clear or test (read) a bit we need to figure out two things:

• which int (of the array) is it in
• which of that int's bits are we talking about

Because we're assuming 32 bits per int, we can just divide by 32 (and truncate) to get the array index we want. Dividing by 32 (BITSPERWORD) is the same as shifting to the right by 5 (SHIFT). So that's what the a[i>>SHIFT] bit is about. You could also write this as a[i/BITSPERWORD] (and in fact, you'd probably get the same or very similar code assuming your compiler has a reasonable optimizer).

Now that we know which element of a we want, we need to figure out which bit. Really, we want the remainder. We could do this with i%BITSPERWORD, but it turns out that i&MASK is equivalent. This is because BITSPERWORD is a power of 2 (2^5 in this case) and MASK is the bottom 5 bits all set.