Bitwise memmove

What is the best way to implement a bitwise memmove ? The method should take an additional destination and source bit-offset and the count should be in bits too.

I saw that ARM provides a non-standard _membitmove , which does exactly what I need, but I couldn't find its source.

Bind's bitset includes isc_bitstring_copy , but it's not efficient

I'm aware that the C standard library doesn't provide such a method, but I also couldn't find any third-party code providing a similar method.

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As a Highlevel-language provides of the smallest unit as 1 Byte, there won't be a standard function that will offer you this option. Maybe you can look for a third party library which offers such functions, but otherwise you would have to code it your self.

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Assuming "best" means "easiest", you can copy bits one by one. Conceptually, an address of a bit is an object (struct) that has a pointer to a byte in memory and an index of a bit in the byte.

struct pointer_to_bit
{
    uint8_t* p;
    int b;
};

void membitmovebl(
    void *dest,
    const void *src,
    int dest_offset,
    int src_offset,
    size_t nbits)
{
    // Create pointers to bits
    struct pointer_to_bit d = {dest, dest_offset};
    struct pointer_to_bit s = {src, src_offset};

    // Bring the bit offsets to range (0...7)
    d.p += d.b / 8; // replace division by right-shift if bit offset can be negative 
    d.b %= 8; // replace "%=8" by "&=7" if bit offset can be negative
    s.p += s.b / 8;
    s.b %= 8;

    // Determine whether it's OK to loop forward
    if (d.p < s.p || d.p == s.p && d.b <= s.b)
    {
        // Copy bits one by one
        for (size_t i = 0; i < nbits; i++)
        {
            // Read 1 bit
            int bit = (*s.p >> s.b) & 1;

            // Write 1 bit
            *d.p &= ~(1 << d.b);
            *d.p |= bit << d.b;

            // Advance pointers
            if (++s.b == 8)
            {
                s.b = 0;
                ++s.p;
            }
            if (++d.b == 8)
            {
                d.b = 0;
                ++d.p;
            }
        }
    }
    else
    {
        // Copy stuff backwards - essentially the same code but ++ replaced by --
    }
}

If you want to write a version optimized for speed, you will have to do copying by bytes (or, better, words), unroll loops, and handle a number of special cases ( memmove does that; you will have to do more because your function is more complicated).

PS Oh, seeing that you call isc_bitstring_copy inefficient, you probably want the speed optimization. You can use the following idea:

Start copying bits individually until the destination is byte-aligned ( db == 0 ). Then, it is easy to copy 8 bits at once, doing some bit twiddling. Do this until there are less than 8 bits left to copy; then continue copying bits one by one.

// Copy 8 bits from s to d and advance pointers
*d.p = *s.p++ >> s.b;
*d.p++ |= *s.p << (8 - s.b);

PPS Oh, and seeing your comment on what you are going to use the code for, you don't really need to implement all the versions (byte/halfword/word, big/little-endian); you only want the easiest one - the one working with words ( uint32_t ).

---

Here is a partial implementation (not tested). There are obvious efficiency and usability improvements.

Copy n bytes from src to dest (not overlapping src ), and shift bits at dest rightwards by bit bits, 0 <= bit <= 7. This assumes that the least significant bits are at the right of the bytes

void memcpy_with_bitshift(unsigned char *dest, unsigned char *src, size_t n, int bit)
{
  int i;

  memcpy(dest, src, n);

  for (i = 0; i < n; i++) {
    dest[i] >> bit;
  }

  for (i = 0; i < n; i++) {
    dest[i+1] |= (src[i] << (8 - bit));
  }
}

Some improvements to be made:

Don't overwrite first bit bits at beginning of dest .

Merge loops

Have a way to copy a number of bits not divisible by 8

Fix for >8 bits in a char

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