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 the solution it was like this:

#define BITSPERWORD 32
#define SHIFT 5
#define MASK 0x1F
#define N 10000000

int a[1 + N/BITSPERWORD]; 

void set(int i) { a[i >> SHIFT] |= (1 << (i & MASK)); 

Where I am getting confused at is this statement

 1 << (i & MASK)

Could someone please explain me what's going on here?

Answer 1

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.

Answer 2

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.