解包十六进制编码的浮点数

Question

我正在尝试将以下 Python 代码翻译成 C++：

import struct
import binascii


inputstring = ("0000003F" "0000803F" "AD10753F" "00000080")
num_vals = 4

for i in range(num_vals):
    rawhex = inputstring[i*8:(i*8)+8]

    # <f for little endian float
    val = struct.unpack("<f", binascii.unhexlify(rawhex))[0]
    print val

    # Output:
    # 0.5
    # 1.0
    # 0.957285702229
    # -0.0

因此它读取 32 位的十六进制编码字符串，使用 unhexlify 方法将其转换为字节数组，并将其解释为 little-endian 浮点值。

以下几乎可以工作，但代码有点糟糕（最后一个 00000080 解析不正确）：

#include <sstream>
#include <iostream>


int main()
{
    // The hex-encoded string, and number of values are loaded from a file.
    // The num_vals might be wrong, so some basic error checking is needed.
    std::string inputstring = "0000003F" "0000803F" "AD10753F" "00000080";
    int num_vals = 4;


    std::istringstream ss(inputstring);

    for(unsigned int i = 0; i < num_vals; ++i)
    {
        char rawhex[8];

// The ifdef is wrong. It is not the way to detect endianness (it's
// always defined)
#ifdef BIG_ENDIAN
        rawhex[6] = ss.get();
        rawhex[7] = ss.get();

        rawhex[4] = ss.get();
        rawhex[5] = ss.get();

        rawhex[2] = ss.get();
        rawhex[3] = ss.get();

        rawhex[0] = ss.get();
        rawhex[1] = ss.get();
#else
        rawhex[0] = ss.get();
        rawhex[1] = ss.get();

        rawhex[2] = ss.get();
        rawhex[3] = ss.get();

        rawhex[4] = ss.get();
        rawhex[5] = ss.get();

        rawhex[6] = ss.get();
        rawhex[7] = ss.get();
#endif

        if(ss.good())
        {
            std::stringstream convert;
            convert << std::hex << rawhex;
            int32_t val;
            convert >> val;

            std::cerr << (*(float*)(&val)) << "\n";
        }
        else
        {
            std::ostringstream os;
            os << "Not enough values in LUT data. Found " << i;
            os << ". Expected " << num_vals;
            std::cerr << os.str() << std::endl;
            throw std::exception();
        }
    }
}

（在 OS X 10.7/gcc-4.2.1 上编译，带有简单的g++ blah.cpp）

特别是，我想摆脱 BIG_ENDIAN 宏的东西，因为我确信有更好的方法来做到这一点，正如 this post 所讨论的那样。

很少有其他随机细节 - 我不能使用 Boost（项目的依赖关系太大）。该字符串通常包含 1536 (8³*3) 到 98304 个浮点值 (32³*3)，最多 786432 (64³*3)

（edit2：添加了另一个值，00000080 == -0.0）

Answer 1

我认为整个istringstring 业务有点矫枉过正。自己解析一个数字要容易得多。

首先，创建一个将十六进制数字转换为整数的函数：

signed char htod(char c)
{
  c = tolower(c);
  if(isdigit(c))
    return c - '0';

  if(c >= 'a' && c <= 'f')
    return c - 'a' + 10;

  return -1;
}

然后简单地将字符串转换为整数。下面的代码不检查错误并假定大端顺序 - 但您应该能够填写详细信息。

unsigned long t = 0;
for(int i = 0; i < s.length(); ++i)
  t |= (t << 4) & htod(s[i]);

那么你的浮动是

float f = * (float *) &t;

Answer 2

以下是为删除#ifdef BIG_ENDIAN 块而修改的更新代码。它使用应该与主机字节顺序无关的读取技术。它通过将十六进制字节（源字符串中的小端）读取为与 iostream std::hex 运算符兼容的大端字符串格式来实现这一点。一旦采用这种格式，主机字节顺序是什么就无关紧要了。

此外，它还修复了 rawhex 在某些情况下需要以零结尾才能插入到 convert 中而没有尾随垃圾的错误。

我没有要测试的大端系统，所以请在您的平台上进行验证。这是在 Cygwin 下编译和测试的。

#include <sstream>
#include <iostream>

int main()
{
    // The hex-encoded string, and number of values are loaded from a file.
    // The num_vals might be wrong, so some basic error checking is needed.
    std::string inputstring = "0000003F0000803FAD10753F00000080";
    int num_vals = 4;
    std::istringstream ss(inputstring);
    size_t const k_DataSize = sizeof(float);
    size_t const k_HexOctetLen = 2;

    for (uint32_t i = 0; i < num_vals; ++i)
    {
        char rawhex[k_DataSize * k_HexOctetLen + 1];

        // read little endian string into memory array
        for (uint32_t j=k_DataSize; (j > 0) && ss.good(); --j)
        {
            ss.read(rawhex + ((j-1) * k_HexOctetLen), k_HexOctetLen);
        }

        // terminate the string (needed for safe conversion)
        rawhex[k_DataSize * k_HexOctetLen] = 0;

        if (ss.good())
        {
            std::stringstream convert;
            convert << std::hex << rawhex;
            uint32_t val;
            convert >> val;

            std::cerr << (*(float*)(&val)) << "\n";
        }
        else
        {
            std::ostringstream os;
            os << "Not enough values in LUT data. Found " << i;
            os << ". Expected " << num_vals;
            std::cerr << os.str() << std::endl;
            throw std::exception();
        }
    }
}

Answer 3

这就是我们最终的结果，OpenColorIO/src/core/FileFormatIridasLook.cpp

（Amardeep 对未签名的uint32_t 修复的回答也可能有效）

    // convert hex ascii to int
    // return true on success, false on failure
    bool hexasciitoint(char& ival, char character)
    {
        if(character>=48 && character<=57) // [0-9]
        {
            ival = static_cast<char>(character-48);
            return true;
        }
        else if(character>=65 && character<=70) // [A-F]
        {
            ival = static_cast<char>(10+character-65);
            return true;
        }
        else if(character>=97 && character<=102) // [a-f]
        {
            ival = static_cast<char>(10+character-97);
            return true;
        }

        ival = 0;
        return false;
    }

    // convert array of 8 hex ascii to f32
    // The input hexascii is required to be a little-endian representation
    // as used in the iridas file format
    // "AD10753F" -> 0.9572857022285461f on ALL architectures

    bool hexasciitofloat(float& fval, const char * ascii)
    {
        // Convert all ASCII numbers to their numerical representations
        char asciinums[8];
        for(unsigned int i=0; i<8; ++i)
        {
            if(!hexasciitoint(asciinums[i], ascii[i]))
            {
                return false;
            }
        }

        unsigned char * fvalbytes = reinterpret_cast<unsigned char *>(&fval);

#if OCIO_LITTLE_ENDIAN
        // Since incoming values are little endian, and we're on little endian
        // preserve the byte order
        fvalbytes[0] = (unsigned char) (asciinums[1] | (asciinums[0] << 4));
        fvalbytes[1] = (unsigned char) (asciinums[3] | (asciinums[2] << 4));
        fvalbytes[2] = (unsigned char) (asciinums[5] | (asciinums[4] << 4));
        fvalbytes[3] = (unsigned char) (asciinums[7] | (asciinums[6] << 4));
#else
        // Since incoming values are little endian, and we're on big endian
        // flip the byte order
        fvalbytes[3] = (unsigned char) (asciinums[1] | (asciinums[0] << 4));
        fvalbytes[2] = (unsigned char) (asciinums[3] | (asciinums[2] << 4));
        fvalbytes[1] = (unsigned char) (asciinums[5] | (asciinums[4] << 4));
        fvalbytes[0] = (unsigned char) (asciinums[7] | (asciinums[6] << 4));
#endif
        return true;
    }