为什么Double.NaN == Double.NaN返回false？ [英] Why does Double.NaN==Double.NaN return false?

问题描述

我只是在研究OCPJP的问题，我发现这个奇怪的代码：
$ b $ pre $ public static void main（String a []） {
System.out.println（Double.NaN == Double.NaN）;
System.out.println（Double.NaN！= Double.NaN）;
}


当我运行代码时，我得到了：

  false 
 true 
  

当我们比较两个看起来相同的东西时，输出如何 false ？ NaN的意思是不是一个数字（NaN）意思是不是一个数字 。

Java语言规范（JLS）第三版说：
$ b

一个溢出的操作产生一个有符号的无穷大，一个下溢的操作产生非规格化的值或有符号的零，并且没有数学上确定的结果的操作产生NaN。以NaN作为操作数的所有数字操作都会产生NaN。正如已经描述的那样，NaN是无序的，所以涉及一个或两个NaN的数字比较操作返回 false 和任何！= 比较涉及NaN返回 true ，包括 x！= x 当 x 是NaN。

I was just studying OCPJP questions and I found this strange code:

public static void main(String a[]) {
    System.out.println(Double.NaN==Double.NaN);
    System.out.println(Double.NaN!=Double.NaN);
}

When I ran the code, I got:

false
true

How is the output false when we're comparing two things that look the same as each other? What does NaN mean?

解决方案

NaN means "Not a Number".

Java Language Specification (JLS) Third Edition says:

An operation that overflows produces a signed infinity, an operation that underflows produces a denormalized value or a signed zero, and an operation that has no mathematically definite result produces NaN. All numeric operations with NaN as an operand produce NaN as a result. As has already been described, NaN is unordered, so a numeric comparison operation involving one or two NaNs returns false and any != comparison involving NaN returns true, including x!=x when x is NaN.

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