在 C++ 中将 int 转换为 16 位浮点数（半精度浮点数）

Question

如何将integer 转换为half precision float（将存储到数组unsigned char[2] 中）。输入 int 的范围为 1-65535。精度真的不是问题。

我正在做类似的事情来将16bit int 转换为unsigned char[2]，但我知道没有half precision float C++ 数据类型。以下示例：

int16_t position16int = (int16_t)data;
memcpy(&dataArray, &position16int, 2);

Answer 1

这是一件非常简单的事情，您需要的所有信息都在Wikipedia。

示例实现：

#include <stdio.h>

unsigned int2hfloat(int x)
{
  unsigned sign = x < 0;
  unsigned absx = ((unsigned)x ^ -sign) + sign; // safe abs(x)
  unsigned tmp = absx, manbits = 0;
  int exp = 0, truncated = 0;

  // calculate the number of bits needed for the mantissa
  while (tmp)
  {
    tmp >>= 1;
    manbits++;
  }

  // half-precision floats have 11 bits in the mantissa.
  // truncate the excess or insert the lacking 0s until there are 11.
  if (manbits)
  {
    exp = 10; // exp bias because 1.0 is at bit position 10
    while (manbits > 11)
    {
      truncated |= absx & 1;
      absx >>= 1;
      manbits--;
      exp++;
    }
    while (manbits < 11)
    {
      absx <<= 1;
      manbits++;
      exp--;
    }
  }

  if (exp + truncated > 15)
  {
    // absx was too big, force it to +/- infinity
    exp = 31; // special infinity value
    absx = 0;
  }
  else if (manbits)
  {
    // normal case, absx > 0
    exp += 15; // bias the exponent
  }

  return (sign << 15) | ((unsigned)exp << 10) | (absx & ((1u<<10)-1));
}

int main(void)
{
  printf(" 0: 0x%04X\n", int2hfloat(0));
  printf("-1: 0x%04X\n", int2hfloat(-1));
  printf("+1: 0x%04X\n", int2hfloat(+1));
  printf("-2: 0x%04X\n", int2hfloat(-2));
  printf("+2: 0x%04X\n", int2hfloat(+2));
  printf("-3: 0x%04X\n", int2hfloat(-3));
  printf("+3: 0x%04X\n", int2hfloat(+3));
  printf("-2047: 0x%04X\n", int2hfloat(-2047));
  printf("+2047: 0x%04X\n", int2hfloat(+2047));
  printf("-2048: 0x%04X\n", int2hfloat(-2048));
  printf("+2048: 0x%04X\n", int2hfloat(+2048));
  printf("-2049: 0x%04X\n", int2hfloat(-2049)); // first inexact integer
  printf("+2049: 0x%04X\n", int2hfloat(+2049));
  printf("-2050: 0x%04X\n", int2hfloat(-2050));
  printf("+2050: 0x%04X\n", int2hfloat(+2050));
  printf("-32752: 0x%04X\n", int2hfloat(-32752));
  printf("+32752: 0x%04X\n", int2hfloat(+32752));
  printf("-32768: 0x%04X\n", int2hfloat(-32768));
  printf("+32768: 0x%04X\n", int2hfloat(+32768));
  printf("-65504: 0x%04X\n", int2hfloat(-65504)); // legal maximum
  printf("+65504: 0x%04X\n", int2hfloat(+65504));
  printf("-65505: 0x%04X\n", int2hfloat(-65505)); // infinity from here on
  printf("+65505: 0x%04X\n", int2hfloat(+65505));
  printf("-65535: 0x%04X\n", int2hfloat(-65535));
  printf("+65535: 0x%04X\n", int2hfloat(+65535));
  return 0;
}

输出（ideone）：

 0: 0x0000
-1: 0xBC00
+1: 0x3C00
-2: 0xC000
+2: 0x4000
-3: 0xC200
+3: 0x4200
-2047: 0xE7FF
+2047: 0x67FF
-2048: 0xE800
+2048: 0x6800
-2049: 0xE800
+2049: 0x6800
-2050: 0xE801
+2050: 0x6801
-32752: 0xF7FF
+32752: 0x77FF
-32768: 0xF800
+32768: 0x7800
-65504: 0xFBFF
+65504: 0x7BFF
-65505: 0xFC00
+65505: 0x7C00
-65535: 0xFC00
+65535: 0x7C00

Answer 2

我问了如何将32位浮点数转换为16位浮点数的问题。

Float32 to Float16

因此，您可以很容易地将 int 转换为浮点数，然后使用上面的问题创建一个 16 位浮点数。我建议这可能比从 int 直接转到 16 位浮点数要容易得多。实际上，通过转换为 32 位浮点数，您已经完成了大部分工作，然后您只需要移动几位即可。

编辑：查看 Alexey 的出色答案，我认为使用硬件 int 进行浮点转换然后对其进行位移很可能比他的方法快一点。可能值得对这两种方法进行分析并进行比较。

Answer 3

在@kbok 问题评论之后，我使用this answer 的第一部分来获取半浮点数，然后获取数组：

uint16_t position16float = float_to_half_branch(data);
memcpy(&dataArray, &position16float, 2);

Answer 4

如果您的目标是支持的硬件，您可以使用内在函数：

https://software.intel.com/en-us/articles/performance-benefits-of-half-precision-floats https://software.intel.com/sites/products/documentation/doclib/iss/2013/compiler/cpp-lin/GUID-7679FF37-257B-4F90-8668-5B3AA62587AD.htm