javascript - JavaScript 运算符 === 是否可证明是可传递的？

Question

JavaScript 古怪的弱类型==运算符可以很容易地显示为不可传递的，如下所示：

var a = "16";
var b = 16;
var c = "0x10";
alert(a == b && b == c && a != c); // alerts true

我想知道是否有任何类似的技巧可以使用舍入误差，Infinity或者NaN应该显示===为非传递性的，或者是否可以证明它确实是传递性的。

Answer 1

Javascript 中的===运算符似乎具有传递性。

NaN可靠地不同于NaN：

>>> 0/0 === 0/0
false
>>> 0/0 !== 0/0
true

Infinity可靠地等于Infinity：

>>> 1/0 === 1/0
true
>>> 1/0 !== 1/0
false

对象（哈希）总是不同的：

>>> var a = {}, b = {};
>>> a === b
false
>>> a !== b
true

And since the === operator does not perform any type coercion, no value conversion can occur, so the equality / inequality semantics of primitive types will remain consistent (i.e. won't contradict one another), interpreter bugs notwithstanding.

Answer 2

如果您查看规范（http://bclary.com/2004/11/07/#a-11.9.6），您会发现没有进行类型强制。此外，其他一切都非常简单，所以也许只有实现错误才会使其不可传递。

Answer 3

Assuming you have variable a,b, and c you can't be 100% certain as shown here. If someone is doing something as hackish as above in production, well then you probably have bigger problems ;)

var a = 1;
var b = 1;
Object.defineProperty(window,"c",{get:function(){return b++;}});

alert(a === b && b === c && a !== c); // alerts true