Question

我在一个几乎没有双/数学库（NETMF）的环境中工作。我写这篇课程是为了让事情变得更轻松：

public struct DoubleEx
{
    public const double NaN = 0.0d / 0.0d;
    public static bool IsNaN(double x)
    {
        return x != x;
    }
    ...
}

好像应该有用，对吧？

好吧，当我运行此代码时：

Debug.Print("Method call: " + DoubleEx.IsNaN(DoubleEx.NaN));
Debug.Print("Method call: " + DoubleEx.NaN != DoubleEx.NaN);

我得到了这个输出：

False
True

不知何故，把它放在一个函数中的行为打破了它！这里有什么样的优化吗？或硬件是否误解了指令？

Answer 1

以下内容基于IEEE标准754：

// @struct IEEE_DOUBLEREP | allows bit access to 8 byte floats
//[StructLayout(LayoutKind.Sequential)]
//public struct ieee_doublerep
//{
//    ulong low_mantissa;       // @field low 16 bits of mantissa
//    ushort mid_mantissa;  // @field mid 16 bits of mantissa
//    uint high_mantissa:4;     // @field high 4 bits of mantissa
//    uint exponent:11;         // @field exponent of floating point number
//    uint sign:1;              // @field sign of floating point number
//};

public struct DoubleEx
{
    public const long NANMASK = 0x7FF0000000000000;
    public const long INFINITYMASK = 0x000FFFFFFFFFFFFF;

    public const double NaN = 0.0f / 0.0f;
    public const double NegativeInfinity = -1.0f / 0.0f;
    public const double PositiveInfinity = 1.0f / 0.0f;
    public static bool IsNaNBad(double x)
    {
        return x != x;
    }

    public unsafe static bool IsNaN(double value)        
    {
        long rep = *((long*)&value);
        return ((rep & NANMASK) == NANMASK &&
                ((rep & INFINITYMASK) != 0));
    }

    public unsafe static bool IsPositiveInfinity(double value)
    {
        double negInf = DoubleEx.PositiveInfinity;
        return *((long*)&value) == *((long*)&negInf);
    }

    public unsafe static bool IsNegativeInfinity(double value)
    {
        double posInf = DoubleEx.PositiveInfinity;
        return *((long*)&value) == *((long*)&posInf);
    }

    public unsafe static bool IsInfinite(double x)
    {
        long rep = *((long*)&x);
        return ((rep & NANMASK) == NANMASK &&
                ((rep & INFINITYMASK) == 0));
    }
}

Answer 2

您有一个运算符优先级问题，并将其置于函数内会更改表达式。

尝试：

Debug.Print("Method call: " + (DoubleEx.NaN != DoubleEx.NaN));

那怎么样：

static bool DoubleInequal(double a, double b) { return a != b; }
static bool IsNaN(double x) { return DoubleInequal(x, x + 0.0); }

Answer 3

你试过return x == NaN吗？假设x！= x与“IsNaN”同义，对我来说似乎不是一种好习惯。

是什么导致这种行为奇怪？

3 个答案: