c - 填充八位字节字符串

Question

我有 65 个不同位长的参数，我需要填写一个八位字节字符串。参数将在八位字节字符串中连续填充。例如，假设第一个参数是 1 位长，因此它将在八位字节字符串的第一个八位字节的第 0 位位置填充。现在第二个参数假设为 9 位长。因此，该参数的前 7 位将被填充到相同的八位字节中，接下来的 2 位应位于下一个八位字节的第 0 位和第 1 位位置。类似地，其他参数将被填充到八位字节字符串中。我试图编写一个函数，其中我将指针传递给当前八位字节、位位置和源指针，数据将从那里复制。但我发现逻辑实现有困难。我尝试了许多逻辑（位操作、位移、旋转等），但无法获得正确的逻辑。如果有人可以在“C”中给我一个逻辑/功能，我将不胜感激。您也可以使用不同的函数原型。

我编写了一个 16 位的代码，如下所示：

    void set16BitVal(U8** p_buf, U8* bitPos, U16 src)
{
        U16 ctr;
        U16 bitVal;
        U16 srcBitVal;
        U16 tempSrc = src;
        U8 temp = **p_buf;
        printf("\n temp = %d\n", temp);
        for(ctr=0; ctr<16; ctr++)
        {
         bitVal = 1;
         bitVal = bitVal << ctr;
         srcBitVal = src & bitVal;

      temp = temp | srcBitVal;
      **p_buf = temp;

      if(srcBitVal)
        srcBitVal = 1;
      else
        srcBitVal = 0;

      printf("\n bit = %d, p_buf = %x \t p_buf=%d  bitPos=%d ctr=%d srcBitVal = %d\n",\
              tempSrc, *p_buf, **p_buf, *bitPos, ctr, srcBitVal);
      *bitPos = (*bitPos+1)%8; /*wrap around after bitPos:7 */
      if(0 == *bitPos)
      {
        (*p_buf)++; /*jump to next octet*/
        temp = **p_buf;
        printf("\n Value of temp = %d\n", temp);
      }



      //printf("\n ctr=%d srcBitVal = %d", ctr, srcBitVal);
      printf("\n");
    }
}

但问题是假设如果我通过 src=54647，我会得到以下输出：

温度 = 0

位 = 54647，p_buf = bf84da4b p_buf = 1 位位置 = 0 ctr = 0 srcBitVal = 1

bit = 54647, p_buf = bf84da4b p_buf=3 bitPos=1 ctr=1 srcBitVal = 1

bit = 54647, p_buf = bf84da4b p_buf=7 bitPos=2 ctr=2 srcBitVal = 1

bit = 54647, p_buf = bf84da4b p_buf=7 bitPos=3 ctr=3 srcBitVal = 0

bit = 54647, p_buf = bf84da4b p_buf=23 bitPos=4 ctr=4 srcBitVal = 1

bit = 54647, p_buf = bf84da4b p_buf=55 bitPos=5 ctr=5 srcBitVal = 1

bit = 54647, p_buf = bf84da4b p_buf=119 bitPos=6 ctr=6 srcBitVal = 1

bit = 54647, p_buf = bf84da4b p_buf=119 bitPos=7 ctr=7 srcBitVal = 0

温度值 = 0

位 = 54647，p_buf = bf84da4c p_buf=0 bitPos=0 ctr=8 srcBitVal = 1

位 = 54647，p_buf = bf84da4c p_buf=0 bitPos=1 ctr=9 srcBitVal = 0

bit = 54647, p_buf = bf84da4c p_buf=0 bitPos=2 ctr=10 srcBitVal = 1

位 = 54647, p_buf = bf84da4c p_buf = 0 位位置 = 3 ctr = 11 srcBitVal = 0

bit = 54647, p_buf = bf84da4c p_buf=0 bitPos=4 ctr=12 srcBitVal = 1

bit = 54647, p_buf = bf84da4c p_buf=0 bitPos=5 ctr=13 srcBitVal = 0

位 = 54647，p_buf = bf84da4c p_buf=0 bitPos=6 ctr=14 srcBitVal = 1

位 = 54647，p_buf = bf84da4c p_buf=0 bitPos=7 ctr=15 srcBitVal = 1

温度值 = 0

然而，预期的输出是：下一个字节应该开始填充 src 的第 8 位以上的值。

有人可以帮我解决吗？

Answer 1

你很幸运。因为我喜欢玩弄，所以我专门为你写了一个 BitBuffer 的通用实现。我没有对它进行彻底的测试（例如，不是所有令人讨厌的极端情况），但正如您将看到的，它通过了我添加到下面代码中的简单测试。

#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;
}

该代码并未针对最大性能进行优化，但我想它仍然应该具有不错的性能。任何额外的性能调整都会显着增加代码大小，我想保持代码相当简单。有些人可能会争辩说，使用>> 3代替/ 8和& 0x7代替% 8会带来更好的性能，但是如果你使用一个像样的 C 编译器，那么如果你启用优化，这正是编译器在内部会做的事情，因此我更喜欢让代码更具可读性.

附加说明
当您将指针传递给多字节数据类型时，请注意字节顺序！例如下面的代码

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

在大端机器（PPC CPU）上运行良好，但在小端机器（例如 x86 CPU）上可能无法提供预期的结果。在小端机器上，您必须先交换字节顺序。您可以使用htonsandhtonl为此目的：

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

在大端机器上，htonX函数/宏通常什么都不做，因为该值已经处于“网络字节顺序”（大端）中，而在小端机器上，它们将交换字节顺序。

传递 uint8_t 指针将始终在任一机器上工作，它只有一个字节，因此没有字节顺序。