Structure for an array of bits in C - c

Question

It has come to my attention that there is no builtin structure for a single bit in C. There is (unsigned) char and int, which are 8 bits (one byte), and long which is 64+ bits, and so on (uint64_t, bool...)
I came across this while coding up a huffman tree, and the encodings for certain characters were not necessarily exactly 8 bits long (like 00101), so there was no efficient way to store the encodings. I had to find makeshift solutions such as strings or boolean arrays, but this takes far more memory.
But anyways, my question is more general: is there a good way to store an array of bits, or some sort of user-defined struct? I scoured the web for one but the smallest structure seems to be 8 bits (one byte). I tried things such as int a : 1 but it didn't work. I read about bit fields but they do not simply achieve exactly what I want to do. I know questions have already been asked about this in C++ and if there is a struct for a single bit, but mostly I want to know specifically what would be the most memory-efficient way to store an encoding such as 00101 in C.

Answer 1

If you're mainly interested in accessing a single bit at a time, you can take an array of unsigned char and treat it as a bit array. For example:
unsigned char array[125];
Assuming 8 bits per byte, this can be treated as an array of 1000 bits. The first 16 logically look like this:
---------------------------------------------------------------------------------
byte | 0 | 1 |
---------------------------------------------------------------------------------
bit | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
---------------------------------------------------------------------------------
Let's say you want to work with bit b. You can then do the following:
Read bit b:
value = (array[b/8] & (1 << (b%8)) != 0;
Set bit b:
array[b/8] |= (1 << (b%8));
Clear bit b:
array[b/8] &= ~(1 << (b%8));
Dividing the bit number by 8 gets you the relevant byte. Similarly, mod'ing the bit number by 8 gives you the relevant bit inside of that byte. You then left shift the value 1 by the bit number to give you the necessary bit mask.
While there is integer division and modulus at work here, the dividend is a power of 2 so any decent compiler should replace them with bit shifting/masking.

Answer 2

It has come to my attention that there is no builtin structure for a single bit in C.
That is true, and it makes sense because substantially no machines have bit-addressible memory.
But anyways, my question is more general: is there a good way to store
an array of bits, or some sort of user-defined struct?
One generally uses an unsigned char or another unsigned integer type, or an array of such. Along with that you need some masking and shifting to set or read the values of individual bits.
I scoured the
web for one but the smallest structure seems to be 8 bits (one byte).
Technically, the smallest addressible storage unit ([[un]signed] char) could be larger than 8 bits, though you're unlikely ever to see that.
I tried things such as int a : 1 but it didn't work. I read about bit
fields but they do not simply achieve exactly what I want to do.
Bit fields can appear only as structure members. A structure object containing such a bitfield will still have a size that is a multiple of the size of a char, so that doesn't map very well onto a bit array or any part of one.
I
know questions have already been asked about this in C++ and if there
is a struct for a single bit, but mostly I want to know specifically
what would be the most memory-efficient way to store an encoding such
as 00101 in C.
If you need a bit pattern and a separate bit count -- such as if some of the bits available in the bit-storage object are not actually significant -- then you need a separate datum for the significant-bit count. If you want a data structure for a small but variable number of bits, then you might go with something along these lines:
struct bit_array_small {
unsigned char bits;
unsigned char num_bits;
};
Of course, you can make that larger by choosing a different data type for the bits member and, maybe, the num_bits member. I'm sure you can see how you might extend the concept to handling arbitrary-length bit arrays if you should happen to need that.

Answer 3

If you really want the most memory efficiency, you can encode the Huffman tree itself as a stream of bits. See, for example:
https://www.siggraph.org/education/materials/HyperGraph/video/mpeg/mpegfaq/huffman_tutorial.html
Then just encode those bits as an array of bytes, with a possible waste of 7 bits.
But that would be a horrible idea. For the structure in memory to be useful, it must be easy to access. You can still do that very efficiently. Let's say you want to encode up to 12-bit codes. Use a 16-bit integer and bitfields:
struct huffcode {
uint16_t length: 4,
value: 12;
}
C will store this as a single 16-bit value, and allow you to access the length and value fields separately. The complete Huffman node would also contain the input code value, and tree pointers (which, if you want further compactness, can be integer indices into an array).

Answer 4

You can make you own bit array in no time.
#define ba_set(ptr, bit) { (ptr)[(bit) >> 3] |= (char)(1 << ((bit) & 7)); }
#define ba_clear(ptr, bit) { (ptr)[(bit) >> 3] &= (char)(~(1 << ((bit) & 7))); }
#define ba_get(ptr, bit) ( ((ptr)[(bit) >> 3] & (char)(1 << ((bit) & 7)) ? 1 : 0 )
#define ba_setbit(ptr, bit, value) { if (value) { ba_set((ptr), (bit)) } else { ba_clear((ptr), (bit)); } }
#define BITARRAY_BITS (120)
int main()
{
char mybits[(BITARRAY_BITS + 7) / 8];
memset(mybits, 0, sizeof(mybits));
ba_setbit(mybits, 33, 1);
if (!ba_get(33))
return 1;
return 0;
};