如何使用CoffeeScript存在操作符来检查某些对象属性是否未定义？ [英] How do I use the CoffeeScript existential operator to check some object properties for undefined?

问题描述

我想使用CoffeeScript存在操作符来检查一些对象属性是否未定义。但是，我遇到了一些问题。

I would like to use the CoffeeScript existential operator to check some object properties for undefined. However, I encountered a little problem.

这样的代码：

console.log test if test?

编译为：

if (typeof test !== "undefined" && test !== null) console.log(test);

这是我想看到的行为。但是，当我尝试使用它对象属性，像这样：

Which is the behavior I would like to see. However, when I try using it against object properties, like this:

console.log test.test if test.test?

我得到类似的结果：

if (test.test != null) console.log(test.test);

这看起来不像是对undefined的检查。我可以通过使用一个更大的检查达到相同（1：1）行为的对象是唯一的方法：

Which desn't look like a check against undefined at all. The only way I could have achieved the same (1:1) behavior as using it for objects was by using a larger check:

console.log test.test if typeof test.test != "undefined" and test.test != null

问题是 - 我做错了什么？或者是编译的代码什么足以检查属性的存在（一个空检查类型转换）？

The question is - am I doing something wrong? Or is the compiled code what is enough to check for existence of a property (a null check with type conversion)?

推荐答案

是与存在操作符的常见混淆点：有时

This is a common point of confusion with the existential operator: Sometimes

x?

编译为

typeof test !== "undefined" && test !== null

而其他时候它只是编译到

and other times it just compiles to

x != null

>这两个是等效的，因为 x！= null 将会是 false > x 是 null 或 undefined 。因此 x！= null 是一种更紧凑的表达方法（x！== undefined&& x！== null）。发生 typeof 编译的原因是编译器认为 x 可能根本没有定义，在这种情况下一个等式测试将触发 ReferenceError：x未定义。

The two are equivalent, because x != null will be false when x is either null or undefined. So x != null is a more compact way of expressing (x !== undefined && x !== null). The reason the typeof compilation occurs is that the compiler thinks x may not have been defined at all, in which case doing an equality test would trigger ReferenceError: x is not defined.

在您的特定情况下， test.test 可能有 undefined 的值，但是你不能得到 ReferenceError 通过引用现有对象上的未定义属性，因此编译器选择较短的输出。

In your particular case, test.test may have the value undefined, but you can't get a ReferenceError by referring to an undefined property on an existing object, so the compiler opts for the shorter output.

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