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位值的十六进制编码字符串，使用unhexlify方法将其转换为字节数组，并将其解释为小端浮点值。
以下几乎可以正常工作，但代码有点蹩脚（最后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可以做到这一点，正如this post讨论的那样。
很少有其他随机细节 - 我不能使用Boost（项目的依赖性太大）。该字符串通常包含1536（8 3 * 3）和98304浮点值（32 3 * 3），最多786432（64 3 * 3）
（edit2：添加了另一个值，00000080 == -0.0）
答案 0 :(得分：1)
我认为整个istringstring业务是一种矫枉过正的行为。一次解析一个数字要容易得多。
首先，创建一个将十六进制数转换为整数的函数：
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。下面的代码不会检查错误并假设有大的字节序 - 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]);
然后你的漂浮
float f = * (float *) &t;
答案 1 :(得分：1)
以下是您修改的更新代码，用于删除#ifdef BIG_ENDIAN块。它使用的读取技术应该与主机字节顺序无关。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。一旦采用这种格式，主机字节顺序无关紧要。
此外，它修复了一个错误，即rawhex需要零终止以插入convert而不会在某些情况下拖尾垃圾。
我没有要测试的大端系统，所以请在您的平台上验证。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();
}
}
}
答案 2 :(得分：-1)
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;
}