在 C++ 中检查双精度(或浮点数)是否为 NaN

Checking if a double (or float) is NaN in C++(在 C++ 中检查双精度(或浮点数)是否为 NaN)
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问题描述

有 isnan() 函数吗?

PS.:我在 MinGW(如果这有所作为).

我通过使用 中的 isnan() 解决了这个问题,它在 中不存在,我是 >#include首先输入.

解决方案

根据 IEEE 标准,NaN 值具有奇怪的特性,即涉及它们的比较总是错误.也就是说,对于一个浮点 f,f != f 将是真的 only 如果 f 是 NaN.

请注意,正如下面的一些评论指出的那样,并非所有编译器在优化代码时都尊重这一点.

对于任何声称使用 IEEE 浮点数的编译器,这个技巧应该有效.但我不能保证它在实践中起作用.如有疑问,请咨询您的编译器.

Is there an isnan() function?

PS.: I'm in MinGW (if that makes a difference).

I had this solved by using isnan() from <math.h>, which doesn't exist in <cmath>, which I was #includeing at first.

解决方案

According to the IEEE standard, NaN values have the odd property that comparisons involving them are always false. That is, for a float f, f != f will be true only if f is NaN.

Note that, as some comments below have pointed out, not all compilers respect this when optimizing code.

For any compiler which claims to use IEEE floating point, this trick should work. But I can't guarantee that it will work in practice. Check with your compiler, if in doubt.

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