Question

我正在尝试编写c函数ReadInt32，WriteInt32，用于在流缓冲区中按顺序打包位，并考虑到大小优化。但是我的函数没有按预期工作，我在缓冲区中写入后找不到相同的值。我需要一些帮助来指出我误解的地方和地点。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define BITSET(x,n)    (x | ( 1 << n))
#define BITCLEAR(x,n)  (x & ~(1 << n))
#define ISBITSET(x,n)  ((x & (1 << n)) != 0)

/**
 @param nbits The number of bits to write
 @param val the integer value to write
 @param bufptr a pointer on the buffer to write in
 @param nbitswritten the number of bits already written in the previous write.
 @return the number of bits actually written.
*/
int WriteInt32(int nbits, int val, uint8_t* bufptr, int nbitswritten)
{
   uint8_t* p=(uint8_t*)&val;
   uint8_t* ptr = &(*(bufptr+nbitswritten));

   for (int i=0; i<nbits; i++)
   {
       int bpos = (i&0x7);
       *(ptr+(i>>3)) = ISBITSET(*(p+(i>>3)), bpos) ? (BITSET(*(ptr+(i>>3)), bpos)) : (BITCLEAR(*(ptr+(i>>3)), bpos));

   }
   return (nbitswritten + nbits);
}

int ReadInt32(int nbits, int& val, uint8_t* ptr, int nbitsread)
{
    val = 0;
    uint8_t* p = &(*(ptr+nbitsread));
    for (int i=0; i<nbits; i++)
    {
        uint8_t ch = *(p+(i>>3));
        int bpos = (i&0x7);
        val = (ISBITSET(ch, bpos) ? BITSET(val, bpos) : BITCLEAR(val, bpos));
    }
    return (nbitsread+nbits);
}

int main (int argc, const char * argv[])
{
    int value = 289;     // for example I want to encode this value on 10 bits
    unsigned char buf[50];

    // packing
    int nbitswritten = 0;
    nbitswritten = WriteInt32 (10, value, buf, 50, nbitswritten);

    // unpacking - read from buffer
    int nbitsread = 0;
    int rvalue;
    nbitsread = ReadInt32(10, rvalue, buf, nbitsread);

    if ( value == rvalue)
        printf("encoding & decoding ok\n");
    else
        printf("encoding or decoding failed\n");

   return 0;
};

谢谢奥利弗

Answer 1

这是我自己的问题的答案。如何写/读一个编码小于其自然大小的整数。如果它可以帮助某人。非常好。 ;）

为例：

struct dd
{  
    short   var1:5;   // This variable is encoded on 5 bits
    int     var2:11;  // This variable is encoded on 11 bits
    uint8_t var3:3;   // This variable is encoded on 3 bits 
};

uint8_t buffer[100];
bzero(buffer, 100);
int nbitswritten = 0;
struct dd vardd = { 16, 452, 3 };

// write stuff
nbitswritten = WriteBits(5, vardd.var1, buffer, 100, nbitswritten);
nbitswritten = WriteBits(11, vardd.var2, buffer, 100, nbitswritten);
nbitswritten = WriteBits(3, vardd.var3, buffer, 100, nbitswritten);

// read back
int v1, v2, v3;
int nbitsread = 0;
struct dd result;
nbitsread = ReadInt(5, v1, buffer, 100, nbitsread);
nbitsread = ReadInt(11, v2, buffer, 100, nbitsread);
nbitsread = ReadInt(3, v3, buffer, 100, nbitsread);

result.var1 = (short)v1;
result.var2 = v2;
result.var3 = (uint8_t)v3;

实际上，Read功能来自Torque Engine库！这是我稍加修改的版本：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define BITSET(x,n)    (x | ( 1 << n))
#define BITCLEAR(x,n)  (x & ~(1 << n))
#define ISBITSET(x,n)  ((x & (1 << n)) != 0)

/**
 @param nbits The number of bits to write
 @param val the value to write
 @param bufptr the buffer to write to/in
 @param buflen the maximum buffer length in byte/octet
 @param nbitswritten the number of bits in bufptr already written.
 @return the number of bits written;
*/
int WriteBits(int nbits, const void * val, uint8_t * bufptr, int buflen, int nbitswritten)
{
    if (!bufptr || !val || !nbits || !buflen)
       return 0;

    if ( nbitswritten >= buflen<<8L )
       return 0;

    if ( nbits+nbitswritten>buflen<<8L )
       return 0;

    uint8_t* p=(uint8_t*)val;
    int n = nbitswritten;
    for (int i=0; i<nbits; i++)
    {
       int sbpos = (i&0x7);
       int dbpos = (n&0x7);
       *(bufptr+(n>>3)) = ISBITSET(*(p+(i>>3)), sbpos) ? (BITSET(*(bufptr+(n>>3)), dbpos)) : (BITCLEAR(*(bufptr+(n>>3)), dbpos));
       n++;
    }
    nbitswritten = n;
    return (n);
}

int ReadBits(int nbits, void * val, uint8_t * bufptr, int buflen, int nbitsread)
{
    if (!nbits)
        return 0;

    uint8_t * stPtr = (bufptr + (nbitsread >> 3));
    int byteCount = (nbits + 7) >> 3;

    uint8_t * ptr = (TUint8*)val;

    int downShift = nbitsread & 0x7;
    int upShift = 8 - downShift;

    uint8_t curB = *stPtr;
    const uint8_t *stEnd = bufptr + buflen;
    while(byteCount--)
    {
       stPtr++;
       uint8_t nextB = stPtr < stEnd ? *stPtr : 0;
       *ptr++ = (curB >> downShift) | (nextB << upShift);
       curB = nextB;
    }

    nbitsread += nbits;
    return nbitsread;
}

int ReadInt(int nbits, int &val, uint8_t * bufptr, int buflen, int nbitsread)
{
    int ret = ReadBits(nbits, &val, bufptr, buflen, nbitsread);
    if (nbits != 32) {
       val &= (1<<nbits)-1;
    }
    return ret;
}

橄榄

C中的可变长度位填料

1 个答案: