为什么Javascript负数不总是真或假？ [英] why are Javascript negative numbers not always true or false?

问题描述

-1 == true;        //false
-1 == false        //false
-1 ? true : false; //true

任何人都可以解释上述输出吗？我知道我可以通过比较0来解决这个问题，但我很感兴趣。我希望至少有一个草率的equals语句是真实的，因为它们进行隐式类型转换，我当然没想到三元组会得出完全不同的结果。

Can anyone explain the above output? I know I could work round this by comparing to 0 but I'm interested. I'd expect at least one of the sloppy equals statements to be true as they do implicit type conversion, and I certainly didn't expect the ternary to come up with a totally different result.

推荐答案

在前两种情况下，布尔值被强制转换为数字 - 1表示 true ，0表示 false 。在最后一种情况下，它是一个强制转换为布尔值的数字，除0和NaN之外的任何数字都将转换为 true 。所以你的测试用例更像是这样：

In the first two cases, the boolean is cast to a number - 1 for true and 0 for false. In the final case, it is a number that is cast to a boolean and any number except for 0 and NaN will cast to true. So your test cases are really more like this:

-1 == 1; // false
-1 == 0; // false
true ? true : false; // true

任何非0或1的数字都是如此。

The same would be true of any number that isn't 0 or 1.

有关更多详细信息，请阅读ECMAScript文档。来自第3版[PDF] ，部分 11.9.3抽象等式比较算法：

For more detail, read the ECMAScript documentation. From the 3rd edition [PDF], section 11.9.3 The Abstract Equality Comparison Algorithm:

19 。如果Type（y）是布尔值，则返回比较结果x == ToNumber（y）。

19. If Type(y) is Boolean, return the result of the comparison x == ToNumber(y).

值得给出全部算法读取，因为其他类型可能导致更糟糕的陷阱。

It's worth giving the full algorithm a read because other types can cause worse gotchas.

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