I noticed that too. The author is probably confusing the fact that you can compute the modulo for a power of two by bitwise-and:ing with (the power of two) - 1.

I noticed that too. The author is probably confusing the fact that you can compute the modulo for a power of two by bitwise-and:ing with (the power of two) - 1. I.e. x & 0xffffffff == x % (0xffffffff + 1) (assuming more than 32 bits everywhere).