IEEE754 浮点值的可移植序列化

Question

我最近一直在研究一个需要存储和加载大量数据（包括单精度浮点值）的系统。我决定对整数的网络字节顺序进行标准化，并决定以大端格式存储浮点值，即：

  |-- Byte 0 --| |-- Byte 1 -|  Byte 2   Byte 3
  #      ####### #     ####### ######## ########
Sign     Exponent          Mantissa
 1b    8b, MSB first    23b, MSB first

理想情况下，我想提供htonl() 和ntohl() 之类的函数，因为我已经使用它们来抽取整数，而且我还想以一种尽可能独立于平台的方式来实现它（而假设 float 类型对应于 IEEE754 32 位浮点值）。有没有办法，可能使用ieee754.h，来做到这一点？

我有一个似乎可行的答案，我将在下面发布它，但它似乎非常缓慢且效率低下，我将不胜感激有关如何使其更快和/或更可靠的任何建议.

Answer 1

这是一个可移植的 IEEE 754 写入例程。 它将以 IEEE 754 格式写入一个 double，而不管主机上的浮点表示。

/*
* write a double to a stream in ieee754 format regardless of host
*  encoding.
*  x - number to write
*  fp - the stream
*  bigendian - set to write big bytes first, elee write litle bytes
*              first
*  Returns: 0 or EOF on error
*  Notes: different NaN types and negative zero not preserved.
*         if the number is too big to represent it will become infinity
*         if it is too small to represent it will become zero.
*/
static int fwriteieee754(double x, FILE *fp, int bigendian)
{
    int shift;
    unsigned long sign, exp, hibits, hilong, lowlong;
    double fnorm, significand;
    int expbits = 11;
    int significandbits = 52;

    /* zero (can't handle signed zero) */
    if (x == 0)
    {
        hilong = 0;
        lowlong = 0;
        goto writedata;
    }
    /* infinity */
    if (x > DBL_MAX)
    {
        hilong = 1024 + ((1 << (expbits - 1)) - 1);
        hilong <<= (31 - expbits);
        lowlong = 0;
        goto writedata;
    }
    /* -infinity */
    if (x < -DBL_MAX)
    {
        hilong = 1024 + ((1 << (expbits - 1)) - 1);
        hilong <<= (31 - expbits);
        hilong |= (1 << 31);
        lowlong = 0;
        goto writedata;
    }
    /* NaN - dodgy because many compilers optimise out this test, but
    *there is no portable isnan() */
    if (x != x)
    {
        hilong = 1024 + ((1 << (expbits - 1)) - 1);
        hilong <<= (31 - expbits);
        lowlong = 1234;
        goto writedata;
    }

    /* get the sign */
    if (x < 0) { sign = 1; fnorm = -x; }
    else { sign = 0; fnorm = x; }

    /* get the normalized form of f and track the exponent */
    shift = 0;
    while (fnorm >= 2.0) { fnorm /= 2.0; shift++; }
    while (fnorm < 1.0) { fnorm *= 2.0; shift--; }

    /* check for denormalized numbers */
    if (shift < -1022)
    {
        while (shift < -1022) { fnorm /= 2.0; shift++; }
        shift = -1023;
    }
    /* out of range. Set to infinity */
    else if (shift > 1023)
    {
        hilong = 1024 + ((1 << (expbits - 1)) - 1);
        hilong <<= (31 - expbits);
        hilong |= (sign << 31);
        lowlong = 0;
        goto writedata;
    }
    else
        fnorm = fnorm - 1.0; /* take the significant bit off mantissa */

    /* calculate the integer form of the significand */
    /* hold it in a  double for now */

    significand = fnorm * ((1LL << significandbits) + 0.5f);


    /* get the biased exponent */
    exp = shift + ((1 << (expbits - 1)) - 1); /* shift + bias */

    /* put the data into two longs (for convenience) */
    hibits = (long)(significand / 4294967296);
    hilong = (sign << 31) | (exp << (31 - expbits)) | hibits;
    x = significand - hibits * 4294967296;
    lowlong = (unsigned long)(significand - hibits * 4294967296);

writedata:
    /* write the bytes out to the stream */
    if (bigendian)
    {
        fputc((hilong >> 24) & 0xFF, fp);
        fputc((hilong >> 16) & 0xFF, fp);
        fputc((hilong >> 8) & 0xFF, fp);
        fputc(hilong & 0xFF, fp);

        fputc((lowlong >> 24) & 0xFF, fp);
        fputc((lowlong >> 16) & 0xFF, fp);
        fputc((lowlong >> 8) & 0xFF, fp);
        fputc(lowlong & 0xFF, fp);
    }
    else
    {
        fputc(lowlong & 0xFF, fp);
        fputc((lowlong >> 8) & 0xFF, fp);
        fputc((lowlong >> 16) & 0xFF, fp);
        fputc((lowlong >> 24) & 0xFF, fp);

        fputc(hilong & 0xFF, fp);
        fputc((hilong >> 8) & 0xFF, fp);
        fputc((hilong >> 16) & 0xFF, fp);
        fputc((hilong >> 24) & 0xFF, fp);
    }
    return ferror(fp);
}

Answer 2

简单得多，并且取决于与您相同的假设（即浮点和整数类型具有相同的字节顺序，并且几乎普遍有效-实际上您永远不会遇到系统不正确的地方）：

#include <string.h>

float htonf(float val) {
    uint32_t rep;
    memcpy(&rep, &val, sizeof rep);
    rep = htonl(rep);
    memcpy(&val, &rep, sizeof rep);
    return val;
}

任何相当好的编译器都会优化掉两个memcpy 调用；它们的存在是为了打败过于急切的严格别名优化，因此最终与htonl 一样高效，加上单个函数调用的开销。

Answer 3

正如上面问题中提到的，我有一个解决我的问题的方法，但我并不特别喜欢它，我欢迎其他答案，所以我在这里而不是在问题中发布它。特别是，它似乎很慢，而且我不确定它是否会破坏严格的混叠以及其他潜在问题。

#include <ieee754.h>

float
htonf (float val)
{
  union ieee754_float u;
  float v;
  uint8_t *un = (uint8_t *) &v;

  u.f = val;
  un[0] = (u.ieee.negative << 7) + ((u.ieee.exponent & 0xfe) >> 1);
  un[1] = ((u.ieee.exponent & 0x01) << 7) + ((u.ieee.mantissa & 0x7f0000) >> 16);
  un[2] = (u.ieee.mantissa & 0xff00) >> 8;
  un[3] = (u.ieee.mantissa & 0xff);
  return v;
}

float
ntohf (float val)
{
  union ieee754_float u;
  uint8_t *un = (uint8_t *) &val;

  u.ieee.negative = (un[0] & 0x80) >> 7;
  u.ieee.exponent = (un[0] & 0x7f) << 1;
  u.ieee.exponent += (un[1] & 0x80) >> 7;
  u.ieee.mantissa = (un[1] & 0x7f) << 16;
  u.ieee.mantissa += un[2] << 8;
  u.ieee.mantissa += un[3];

  return u.f;
}