Bit field structure arrays in C - c

Question

How can I create a bit-field array with variable size?
The following code is what I tried, which didn't work.
#include <stdio.h>
int main()
{
int n=4;
struct bite{
unsigned a1:2;
unsigned a2:2;
:
:
unsigned a(n-1):2;
unsigned a(n):2;
}bits;
for(i=1;i<=n;i++)
bits.a[i]=i;
for(i=1;i<=n;i++)
printf("%d ",bits.a[i]);
return 0;
}

Answer 1

The members of a struct cannot be defined at runtime.
You could simulate a bit array using a char array and some macros.
#define BitArray(array, bits) \
unsigned char array[bits / 8 + 1]
#define SetBit(array, n) \
do { array[n / 8] |= (1 << (n % 8)) } while (0)
#define GetBit(array, n) \
((array[n / 8] >> (n % 8)) & 1)
int main(void)
{
BitArray(bits, 42); /* Define 42 bits and init to all 0s
(in fact this allocates memory for (42/8 + 1) * 8 bits). */
SetBit(bits, 2); /* Set bit 2. */
int bit2 = GetBit(bits, 2); /* Get bit 2 */
...
Similar for 2-bit words are per your code:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#define Define2BitWordArray(array, two_bit_words) \
unsigned char array[two_bit_words / 4 + 1]
#define Set2BitWord(array, n, value) \
do { array[n / 4] |= (unsigned char)((value & 0x11) << (2 * (n % 4))); } while (0)
#define Get2BitWord(array, n) \
((array[n / 4] >> (2 * (n % 4))) & 0x11)
int main(void)
{
size_t n = 10;
Define2BitWordArray(bits, n); /* Define 10 two-bits words
(in fact this allocates memory
for (10/4 + 1) * 4 two-bit bits). */
memset(bits, 0, sizeof bits); /* Set all bits to 0. */
for(size_t i = 0; i < n; i++) /* C array's indexes are 0-based. */
{
Set2BitWord(bits, i, i);
}
for(size_t i = 0; i < n; i++)
{
printf("%d ", Get2BitWord(bits, i));
}
}

Answer 2

Remember your array always starts at 0 position.
You can do this:
#include <stdio.h>
typedef struct
{
unsigned int x : 1;
} one;
int main() {
int n = 10;
one xx[n];
int i;
for(i=0;i<n;i++)
xx[i].x=i;
for(i=0;i<n;i++)
printf("%d ",xx[i].x);
return 0;
}

Answer 3

if you want to operate on the particular bit or bits on data which has not limited size (actually limited to the max value of the size_t divided by 8). Little endian in this example
void setbit(void *obj, size_t bit)
{
uint8_t *p = obj;
p[bit >> 3] |= (1 << (bit & 7));
}
void resetbit(void *obj, size_t bit)
{
uint8_t *p = obj;
p[bit >> 3] &= ~(1 << (bit & 7));
}
void assignbit(void *obj, size_t bit, unsigned value)
{
uint8_t *p = obj;
p[bit >> 3] &= ~(1 << (bit & 7));
p[bit >> 3] != (!!value << (bit & 7));
}
void setbits(void *obj, size_t bit, size_t num)
{
while (num--)
setbit(obj, bit + num);
}
void resetbits(void *obj, size_t bit, size_t num)
{
while (num--)
resetbit(obj, bit + num);
}
void assignbits_slow(void *obj, size_t pos, size_t size, unsigned value)
{
for (size_t i = 1, j = 0; j < size; j++, i <<= 1)
assignbit(obj, pos + j, !!(value & i));
}
int getbit(void *obj, size_t bit)
{
uint8_t *p = obj;
return !!(p[bit >> 3] & (1 << (bit & 7)));
}
void *getbits(void *obj, void *buff, size_t bit, size_t nbits)
{
memset(buff, 0, nbits >> 3 + !!(nbits & 7));
do
{
nbits--;
assignbit(buff, bit + nbits, getbit(obj, bit + nbits));
} while (nbits);
return buff;
}