在 float32 中保存 float16 最大数量

Question

如何以 float32 (https://en.wikipedia.org/wiki/Single-precision_floating-point_format) 格式保存 float16 (https://en.wikipedia.org/wiki/Half-precision_floating-point_format) 最大值？

我想要一个可以将 0x7bff 转换为 65504 的函数。0x7bff 是可以用浮点半精度表示的最大值：

0 11110 1111111111 -> decimal value: 65504 

我想让 0x7bff 代表我程序中的实际位。

float fp16_max = bit_cast(0x7bff); 
# want "std::cout << fp16_max" to be 65504

我试图实现这样的功能，但它似乎不起作用：

float bit_cast (uint32_t fp16_bits) {
    float i;
    memcpy(&i, &fp16_bits, 4);
    return i; 
}    
float test = bit_cast(0x7bff);
# print out test: 4.44814e-41

Answer 1

#include <cmath>
#include <cstdio>


/*  Decode the IEEE-754 binary16 encoding into a floating-point value.
    Details of NaNs are not handled.
*/
static float InterpretAsBinary16(unsigned Bits)
{
    //  Extract the fields from the binary16 encoding.
    unsigned SignCode        = Bits >> 15;
    unsigned ExponentCode    = Bits >> 10 & 0x1f;
    unsigned SignificandCode = Bits       & 0x3ff;

    //  Interpret the sign bit.
    float Sign = SignCode ? -1 : +1;

    //  Partition into cases based on exponent code.

    float Significand, Exponent;

    //  An exponent code of all ones denotes infinity or a NaN.
    if (ExponentCode == 0x1f)
        return Sign * (SignificandCode == 0 ? INFINITY : NAN);

    //  An exponent code of all zeros denotes zero or a subnormal.
    else if (ExponentCode == 0)
    {
        /*  Subnormal significands have a leading zero, and the exponent is the
            same as if the exponent code were 1.
        */
        Significand = 0 + SignificandCode * 0x1p-10;
        Exponent    = 1 - 0xf;
    }

    //  Other exponent codes denote normal numbers.
    else
    {
        /*  Normal significands have a leading one, and the exponent is biased
            by 0xf.
        */
        Significand = 1 + SignificandCode * 0x1p-10;
        Exponent    = ExponentCode - 0xf;
    }

    //  Combine the sign, significand, and exponent, and return the result.
    return Sign * std::ldexp(Significand, Exponent);
}


int main(void)
{
    unsigned Bits = 0x7bff;
    std::printf(
        "Interpreting the bits 0x%x as an IEEE-754 binary16 yields %.99g.\n",
        Bits,
        InterpretAsBinary16(Bits));
}

Answer 2

通过 float fp16_max 的声明，您的值已经是 32 位浮点数；无需在这里投射。我想你可以简单地：

float i = fp16_max;

这里的假设是您的“魔术”bit_cast 函数已经正确返回了 32 位浮点数。由于您没有向我们展示 bit-cast 的作用或实际返回的内容，因此我假设它确实返回了正确的 float 值。

Answer 3

如何将float16最大数保存为float32格式？

65504

您可以简单地将整数转换为浮点数：

float half_max = 65504;

如果要计算值，可以使用ldexpf：

float half_max = (2 - ldexpf(1, -10)) * ldexpf(1, 15)

或者一般来说，对于任何 IEEE 浮点数：

// in case of half float
int bits = 16;
int man_bits = 10;

// the calculation
int exp_bits = bits - man_bits - 1;
int exp_max = (1 << (exp_bits - 1)) - 1;
long double max = (2 - ldexp(1, -1 * man_bits)) * ldexp(1, exp_max);

---

位转换 0x7bff 不起作用，因为 0x7bff 是 binary16 格式（在某些字节序中）的表示，而不是 binary32 格式的表示。您不能对相互冲突的表示进行位转换。