将float16最大值保存在float32中

Question

如何将float16 (format)最大值保存为float32 (format)格式？

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

0 11110 1111111111 -> decimal value: 65504 

我希望有0x7bff来表示程序中的实际位数。

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

我试图实现这样一个函数，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

通过声明float fp16_max，您的值已经是32位浮点数；无需在这里进行强制转换。我想你可以简单地：

float i = fp16_max;

这里的假设是，您的“神奇”bit_cast函数已经正确返回了32位浮点数。由于您尚未向我们展示bit-cast所做或实际返回的内容，我将假定它确实返回了一个适当的float值。

Answer 3

如何将float16最大值保存为float32格式？ 65504

您可以简单地将整数转换为浮动：

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)

或者一般情况下，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);

位铸造0x7bff不工作，因为0x7bff是binary16格式的表示形式(在某种程度上)，而不是binary32格式。您不能使用相互冲突的表示。