Float.NaN == Float.NaN [英] Float.NaN == Float.NaN

问题描述

为什么这个比较给我'假'？我看着源代码和Float.NaN定义为

Why does this comparison give me 'false'? I looked at the source and Float.NaN is defined as

/** 
 * A constant holding a Not-a-Number (NaN) value of type
 * <code>float</code>.  It is equivalent to the value returned by
 * <code>Float.intBitsToFloat(0x7fc00000)</code>.
 */
public static final float NaN = 0.0f / 0.0f;

编辑：令人惊讶的是，如果我这样做：

surprisingly, if I do this:

System.out.println("FC " + (Float.compare(Float.NaN, Float.NaN)));

它给我 0 。因此 Float.compare（）确实认为NaN 是等于其本身！

it gives me 0. So Float.compare() does think that NaN is equal to itself!

推荐答案

因为Java实现了IEEE-754浮点标准，保证任何与 NaN 的比较都会返回false（！= ，它返回true）

Because Java implements the IEEE-754 floating point standard which guarantees that any comparison against NaN will return false (except != which returns true)

这意味着你不能以常用的方式检查浮点数是否为NaN，可以重新解释这两个数字作为int和比较或使用更聪明的解决方案：

That means you can't check in your usual ways whether a a floating point number is NaN, so you could either reinterpret both numbers as ints and compare then or use the much cleverer solution:

def isNan(val):
     return val != val

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