为什么我不能.call（）Function.call？ [英] Why can't I .call() Function.call?

问题描述

在Javascript中， Function.call（）可以在给出 this <的情况下调用 Function / code>值和零个或多个参数。

In Javascript, Function.call() can call Function given a this value and zero or more arguments.

Function.call 本身是一个函数。因此，从理论上讲， Function.call 应该与 Function.call.call 相同（或类似地起作用）。

Function.call itself is a function. So in theory, Function.call should be the same (or similarly acting) function as Function.call.call.

在V8中，情况似乎是这样的：

In V8, this seems to be the case:

> Function.call === Function.call.call
true

当我们调用 Function.call（），我们得到一个匿名函数

When we call Function.call(), we get an anonymous function

> Function.call()
[Function: anonymous]

但是，我不能打电话 .call（）在 Function.call 上。

However, I can't call .call() on Function.call.

> Function.call.call()
TypeError: undefined is not a function
at repl:1:21
at REPLServer.defaultEval (repl.js:132:27)
at bound (domain.js:291:14)
at REPLServer.runBound [as eval] (domain.js:304:12)
at REPLServer.<anonymous> (repl.js:279:12)
at REPLServer.emit (events.js:107:17)
at REPLServer.Interface._onLine (readline.js:214:10)
at REPLServer.Interface._line (readline.js:553:8)
at REPLServer.Interface._ttyWrite (readline.js:830:14)
at ReadStream.onkeypress (readline.js:109:10)

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这是怎么回事？ Function.call 显然是一个函数-它不是 undefined ，如错误消息所提示的那样。

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What is going on here? Function.call is clearly a function - it's not undefined as this error message suggests.

推荐答案

简短答案：错误消息是非常令人误解。

Short answer: The error message is very misleading. It is the same error message you get when you do

(undefined)();

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更长的答案：

第二个 .call（）用 this Function.call 中的$ c>。

The second .call() is being invoked with a this of Function.call.

不带参数调用会导致它调用 this 且未定义 作为此值。

Calling it with no parameters causes it to call this with undefined as the this value.

因此，您确实在做

Function.call.call(undefined)

这意味着（隐喻地）您正在做

which means you're (metaphorically) doing

undefined.call()

这实际上只是

undefined()

不传递任何内容（或 undefined ） Function.call.call（）的 this 参数本质上是否定 this 第一个 Function.call（）的上下文（这本身就是 Function 本身），导致在未定义上调用 .call（）。

Passing nothing (or undefined) to the this parameter of Function.call.call() is essentially negating the this context of the first Function.call() (which would be just Function itself), causing .call() to be invoked on undefined.

这将产生错误消息：未定义不是函数。

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