c - Explanation of an algorithm to set, clear and test a single bit

Question

Hey, in the Programming Pearls book, there is a source code for setting, clearing and testing a bit of the given index in an array of ints that is actually a set representation.

The code is the following:

#include<stdio.h>
#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));
}

void clr(int i)
{
    a[i>>SHIFT] &= ~(1<<(i & MASK));
}

int test(int i)
{
    a[i>>SHIFT] & (1<<(i & MASK));
}

Could somebody explain me the reason of the SHIFT and the MASK defines? And what are their purposes in the code?

I've already read the previous related question.

Answer 1

VonC posted a good answer about bitmasks in general. Here's some information that's more specific to the code you posted.

Given an integer representing a bit, we work out which member of the array holds that bit. That is: Bits 0 to 31 live in a[0], bits 32 to 63 live in a[1], etc. All that i>>SHIFT does is i / 32. This works out which member of a the bit lives in. With an optimising compiler, these are probably equivalent.

Obviously, now we've found out which member of a that bitflag lives in, we need to ensure that we set the correct bit in that integer. This is what 1 << i does. However, we need to ensure that we don't try to access the 33rd bit in a 32-bit integer, so the shift operation is constrained by using 1 << (i & 0x1F). The magic here is that 0x1F is 31, so we'll never left-shift the bit represented by i more than 31 places (otherwise it should have gone in the next member of a).