（A == B == C）比较在JavaScript中是如何工作的？ [英] How does (A == B == C) comparison work in JavaScript?

问题描述

我期望下面的比较给出一个错误：

  var A = B = 0; 
 if（A == B == 0）
 console.log（true）; 
 else 
 console.log（false）; 
  

但奇怪的是它会返回 false p>

更奇怪的是，

  console.log == 1））; 
  

返回 true 。

这种三元比较是如何工作的？

解决方案

要理解一个数字和一个布尔值之间的 == 比较将导致布尔值的内部类型转换为一个数字（ true 变为 1 和 false 变为 0

您显示的表达式是从左到右计算的。因此，首先

  A == B 
  

并且结果为 true ，并且您正在将 true 与0 。由于 true 在比较期间变为 1 ，所以 1 == 0 计算结果为 false 。但是当你说

  console.log（（A == B == 1））; 
  

A == B $ c> true ，当与数字比较时，变成 1 ，您再次与1进行比较。这就是为什么它打印 true 。

I was expecting the following comparison to give an error:

var A = B = 0;
if(A == B == 0)
    console.log(true);
else
    console.log(false);

but strangely it returns false.

Even more strangely,

console.log((A == B == 1)); 

returns true.

How does this "ternary" kind of comparison work?

解决方案

First, we need to understand that a == comparison between a number and a boolean value will result in internal type conversion of Boolean value to a number (true becomes 1 and false becomes 0)

The expression you have shown is evaluated from left to right. So, first

A == B

is evaluated and the result is true and you are comparing true with 0. Since true becomes 1 during comparison, 1 == 0 evaluates to false. But when you say

console.log((A == B == 1));

A == B is true, which when compared with number, becomes 1 and you are comparing that with 1 again. That is why it prints true.

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