double.NaN - 这个违反直觉的功能是如何工作的？
我偶然发现了 .NET 对 double.NaN：
public const double NaN = (double)0.0 / (double)0.0;
这在 PositiveInfinity 和 NegativeInfinity 中的做法类似。
double.IsNaN (删除一些#pragmas和评论）被定义为：
[Pure]
[ReliabilityContract(Consistency.WillNotCorruptState, Cer.Success)]
public static bool IsNaN(double d)
{
if (d != d)
{
return true;
}
else
{
return false;
}
}
这对我来说非常违反直觉。
为什么 NaN 定义为除以零？ 0.0 / 0.0 如何表示“幕后”？ double 中如何能除以 0，为什么会 NaN != NaN？
I stumbled upon .NET's definition of double.NaN in code:
public const double NaN = (double)0.0 / (double)0.0;
This is done similarly in PositiveInfinity and NegativeInfinity.
double.IsNaN (with removing a few #pragmas and comments) is defined as:
[Pure]
[ReliabilityContract(Consistency.WillNotCorruptState, Cer.Success)]
public static bool IsNaN(double d)
{
if (d != d)
{
return true;
}
else
{
return false;
}
}
This is very counter-intuitive to me.
Why is NaN defined as division by zero? How is 0.0 / 0.0 represented "behind the scenes"? How can division by 0 be possible in double, and why does NaN != NaN?
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这里的答案相当简单。 .Net框架实现了IEEE (
System.Double符合二进制浮点运算的 IEC 60559:1989 (IEEE 754) 标准。）。这是因为浮点运算实际上必须在许多系统上工作，而不仅仅是 x86/64 架构，因此通过遵循约定，可以确保减少兼容性问题（例如将代码从 DSP 移植到 x86 处理器）。至于d != d，这是一个性能优化。基本上，该指令依赖于硬件指令，该指令可以非常快速地确定两个双浮点数是否相等。根据标准，NAN != NAN 因此是最快的测试方法。试图为您找到一个参考。
Fairly simple answer here. .Net framework has implemented the floating point standard specified by the IEEE (
System.Doublecomplies with the IEC 60559:1989 (IEEE 754) standard for binary floating-point arithmetic.). This is because floating point arithmetic actually has to work across many systems, not just x86/64 architectures, so by following the conventions this ensures that there will be less compatibility issues (for instance porting code from a DSP into an x86 processor).As for the d != d, this is a performance optimisation. Basically this instruction relies on a hardware instruction which can very quickly determine if two double floating point numbers are equal. Under the standard, NAN != NAN and therefore is the fastest way to test. Trying to find a reference for you.
所有这些都是由 IEEE 754 标准强制执行的，几乎所有现代 CPU 都实施该标准。
通过将所有位设置为 1 的指数和至少一位设置为 1 的尾数。请注意，这意味着有大量不同的位模式都表示 NaN - 但是，如上所述，即使位模式相同，必须将它们视为不相等（即 == 必须返回 false）。
All of these are mandated by the IEEE 754 standard, which pretty much all modern CPUs implement.
By having an exponent with all bits set to 1 and a mantissa with at least one bit set to 1. Note that this means that there are a large number of different bit patterns that all represent NaN - but, as mentioned above, even if the bit patterns are identical, they must be considered not equal (i.e. == must return false).
来自 C# 规范：
至于 NaN 在幕后如何表示，请参考 IEEE 的 wikipedia 文章规范有一些例子。
From the C# Spec:
As to how NaN is represented behind the scenes, the wikipedia article on the IEEE spec has some examples.