Why isn't there integer division in this line of C code in the gcc documentation for limits.h?-CodePudding

In limits.h's documentation, it states:

A common reason that a program needs to know how many bits are in an integer type is for using an array of unsigned long int as a bit vector. You can access the bit at index n with:
vector[n / ULONG_WIDTH] & (1UL << (n % ULONG_WIDTH))

But if vector is an unsigned long we'd need n <= ULONG_WIDTH, meaning that the division there would always result in 0. Is vector supposed to be defined instead as an array of unsigned longs or something else?

CodePudding user response：

vector is an unsigned long we'd need n <= ULONG_WIDTH

Vector is an array of unsigned longs.

n is not limited by ULONG_WIDTH. n is the bit position within all the bits in all elements in the array. If you would lay out all bits in all vector elements one after another, n is the bit position within all the bits.

|----------vector[0]----------->|----------vector[1]----------->|----------vector[2]----------->
abcde......................................................................x....................
^LSB                        MSB^ etc.

The bit marked x has position n = 2 * 32 12 = 76, if i'm counting right. To access it, you have to access 12'th bit of 2'nd element of vector. Note: I am indexing from 0, first bit has position 0, first vector element is element number 0.

n / ULONG_WIDTH computes the 2'nd vector element.

n % ULONG_WIDTH computes the 12'th bit position inside 2'nd vector element.

(1UL << (n % ULONG_WIDTH)) is the mask 0b00.....1 with 12'th bit set.

vector[n / ULONG_WIDTH] & (1UL << (n % ULONG_WIDTH)) returns the 12th bit set if 12th bit is set in vector[2], or returns 0.

Typically in the same fashion:
vector[n / ULONG_WIDTH] |= (1UL << (n % ULONG_WIDTH)) set's the bit,
vector[n / ULONG_WIDTH] &= ~(1UL << (n % ULONG_WIDTH)) clears the bit,
vector[n / ULONG_WIDTH] ^= (1UL << (n % ULONG_WIDTH)) toggles the bit.