Bit Mask usage in the program below from Programming Pearls

Question

I started reading \"Programming Pearls\" today and while doing it\'s exercise I came across this question \"How would you implement your own bit vector?\". When I looked at

Answer 1

0x20 is 32, so i & 0x1F takes i modulo 32, so that you never shift by 32 bits. This is a safeguard because shifting by anything that isn't strictly less than the size of the type is undefined behaviour.

Answer 2

Note that MASK is set such that it has the lowest SHIFT bits set, where SHIFT is exactly the base-2 logarithm of BITSPERWORD.

Therefore (i & MASK) will select the lowest 5 bits of i, which is the same as taking the remainder after dividing by 32 (just consider how taking the lowest two digits of a decimal number gives you the remainder after dividing by 100, for example). That gives the number of the bit within a word we're interested in.

1 << (i & MASK)) (which is, by the way, an expression, not a statement) now creates a value where exactly the bit we're interested in is set. Merging this value into the memory word with |= will set the desired bit of the bit vector.