Filling an octet string

Question

I have 65 parameters of different bit length which I need to fill in an octet string. Parametrs would be filled continuously in octet string. For example, suppose the first para

Answer 1

You are lucky. Since I love bit twiddling, I wrote a generic implementation of a BitBuffer just for you. I have not tested it thoroughly (e.g. not all nasty corner cases) but as you will see, it passes the simple tests I added to the code below.

#include <assert.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

struct BitBuffer {
    unsigned length;       // No of bits used in buffer
    unsigned capacity;     // No of bits available in buffer
    uint8_t buffer[];
};

struct BitBuffer * newBitBuffer (
    unsigned capacityInBits
) {
    int capacityInBytes;
    struct BitBuffer * result;

    capacityInBytes = (capacityInBits / 8);
    if (capacityInBits % 8 != 0) {
        capacityInBytes++;
    }

    result = malloc(sizeof(*result) + capacityInBytes);
    if (result) {
        result->length = 0;
        result->capacity = capacityInBits;
    }
    return result;
}

bool addBitsToBuffer (
    struct BitBuffer * bbuffer, const void * bits, unsigned bitCount
) {
    unsigned tmpBuf;
    unsigned tmpBufLen;
    unsigned tmpBufMask;
    uint8_t * nextBufBytePtr;
    const uint8_t * nextBitsBytePtr;

    // Verify input parameters are sane
    if (!bbuffer || !bits) {
        // Evil!
        return false;
    }
    if (bitCount == 0) {
        // No data to add? Nothing to do.
        return true;
    }

    // Verify we have enough space available
    if (bbuffer->length + bitCount > bbuffer->capacity) {
        // Won't fit!
        return false;
    }

    // Get the first byte we start writing bits to
    nextBufBytePtr = bbuffer->buffer + (bbuffer->length / 8);

    // Shortcut:
    // If we happen to be at a byte boundary,
    // we can simply use memcpy and save us a lot of headache.
    if (bbuffer->length % 8 == 0) {
        unsigned byteCount;

        byteCount = bitCount / 8;
        if (bitCount % 8 != 0) {
            byteCount++;
        }
        memcpy(nextBufBytePtr, bits, byteCount);
        bbuffer->length += bitCount;
        return true;
    }

    // Let the bit twiddling begin
    nextBitsBytePtr = bits;
    tmpBuf = *nextBufBytePtr;
    tmpBufLen = bbuffer->length % 8;
    tmpBuf >>= 8 - tmpBufLen;
    tmpBufMask = (~0u) >> ((sizeof(unsigned) * 8) - tmpBufLen);
    // tmpBufMask has the first tmpBufLen bits set to 1.
    // E.g. "tmpBufLen == 3" ==> "tmpBufMask == 0b111 (7)"
    // or "tmpBufLen == 6" ==> "tmpBufMask = 0b111111 (63)", and so on.

    // Beyond this point we will neither access bbuffer->length again, nor
    // can this function fail anymore, so we set the final length already.
    bbuffer->length += bitCount;

    // Process input bits in byte chunks as long as possible
    while (bitCount >= 8) {
        // Add 8 bits to tmpBuf
        tmpBuf = (tmpBuf << 8) | *nextBitsBytePtr;
        // tmpBuf now has "8 + tmpBufLen" bits set

        // Add the highest 8 bits of tmpBuf to our BitBuffer
        *nextBufBytePtr = (uint8_t)(tmpBuf >> tmpBufLen);

        // Cut off the highest 8 bits of tmpBuf
        tmpBuf &= tmpBufMask;
        // tmpBuf now has tmpBufLen bits set again

        // Skip to next input/output byte
        bitCount -= 8;
        nextBufBytePtr++;
        nextBitsBytePtr++;
    }

    // Test if we still have bits left. That will be the case
    // if the input bit count was no integral multiple of 8.
    if (bitCount != 0) {
        // Add bitCount bits to tmpBuf
        tmpBuf = (tmpBuf << bitCount) | (*nextBitsBytePtr >> (8 - bitCount));
        tmpBufLen += bitCount;
    }

    // tmpBufLen is never 0 here, it must have a value in the range [1, 14].
    // We add zero bits to it so that tmpBuf has 16 bits set.
    tmpBuf <<= (16 - tmpBufLen);

    // Now we only need to add one or two more bytes from tmpBuf to our
    // BitBuffer, depending on its length prior to adding the zero bits.
    *nextBufBytePtr = (uint8_t)(tmpBuf >> 8);
    if (tmpBufLen > 8) {
        *(++nextBufBytePtr) = (uint8_t)(tmpBuf & 0xFF);
    }
    return true;
}



int main ()
{
    bool res;
    uint8_t testData[4];
    struct BitBuffer * buf;

    buf = newBitBuffer(1024); // Can hold up to 1024 bits
    assert(buf);

    // Let's add some test data.

    // Add 1 bit "1" => Buffer "1"
    testData[0] = 0xFF;
    res = addBitsToBuffer(buf, testData, 1);
    assert(res);

    // Add 6 bits "0101 01" => Buffer "1010 101"
    testData[0] = 0x54;
    res = addBitsToBuffer(buf, testData, 6);
    assert(res);

    // Add 4 Bits "1100" => Buffer "1010 1011 100"
    testData[0] = 0xC0;
    res = addBitsToBuffer(buf, testData, 4);
    assert(res);

    // Add 16 Bits "0111 1010 0011 0110"
    // ==> Buffer "1010 1011 1000 1111 0100 0110 110
    testData[0] = 0x7A;
    testData[1] = 0x36;
    res = addBitsToBuffer(buf, testData, 16);
    assert(res);

    // Add 5 Bits "0001 1"
    // ==> Buffer "1010 1011 1000 1111 0100 0110 1100 0011"
    testData[0] = 0x18;
    res = addBitsToBuffer(buf, testData, 5);
    assert(res);

    // Buffer should now have exactly a length of 32 bits
    assert(buf->length == 32);

    // And it should have the value 0xAB8F46C3
    testData[0] = 0xAB;
    testData[1] = 0x8F;
    testData[2] = 0x46;
    testData[3] = 0xC3;
    assert(memcmp(buf->buffer, testData, 4) == 0);

    free(buf);
    return 0;
}

The code is not optimized for maximum performance, yet I guess it should have a decent performance nonetheless. Any additional performance tweaks would have increased the code size noticeably and I wanted to keep the code rather simple. Some people may argue that using >> 3 instead of / 8 and & 0x7 instead of % 8 will lead to better performance, yet if you use a decent C compiler, that's exactly what the compiler will internally do anyway if you enable optimizations and thus I preferred to keep the code more readable instead.

Additional Note
When you pass pointers to multi-byte data types, watch the byte order! E.g. the following code

 uint16_t x16 = ...;
 addBitsToBuffer(buf, &x16, ...);
 uint32_t x32 = ...;
 addBitsToBuffer(buf, &x32, ...);

works fine on a big endian machine (PPC CPU), yet it may not give the expected results on a little endian machine (e.g. x86 CPU). On a little endian machine you would have to swap the byte order first. You can use htons and htonl for this purpose:

 uint16_t x16 = ...;
 uint16_t x16be = htons(x16);
 addBitsToBuffer(buf, &x16be, ...);
 uint32_t x32 = ...;
 uint32_t x32be = htonl(x32);
 addBitsToBuffer(buf, &x32be, ...);

On a big endian machine, htonX functions/macros usually do nothing, since the value is already in "network byte order" (big endian), while on a little endian machine they will swap the byte order.

Passing an uint8_t pointer will always work on either machine, it is only a single byte, hence there is no byte order.