为什么名称功能表达式在函数体外不可用 [英] Why a Name Function Expression not available outside function body

问题描述

命名函数表达式，定义为

The Named Function Expression which is defined as

var ninja = function myNinja();

有一种无法解决的行为。

has a behavior which is not able to get through my head.

查看以下代码

 var ninja = function myNinja() {
   console.log(typeof myNinja) //prints 'function'
 };
 console.log(typeof myNinja) //prints 'undefined'

现在， myNinja 是一个命名函数，据我所知，javascript允许命名函数超出其自身函数的范围。

Now, myNinja is a named function and as far as I know javascript allow the named function to go beyond the scope of its own function.

这让我感到困惑。

推荐答案

现在， myNinja 是一个命名函数，据我所知javascript允许命名函数超出其自身函数的范围。

Now, myNinja is a named function and as far as I know javascript allow the named function to go beyond the scope of its own function.

仅在函数声明中。对于命名函数表达式，具体而言不。这就是规范中如何定义的。

Only in a function declaration. It's specifically not the case for a named function expression. It's just how this is defined in the specification.

所有血腥细节都是在规范中，最相关的位是：

All the gory details are in the spec, the most relevant bit is:

注意 FunctionExpression 中的 Identifier 可以从 FunctionExpression的 FunctionBody 中引用，以允许该函数以递归方式调用自身。但是，与 FunctionDeclaration 不同， FunctionExpression 中的标识符无法引用，也不会影响包含 FunctionExpression <的范围/ em>。

NOTE The Identifier in a FunctionExpression can be referenced from inside the FunctionExpression's FunctionBody to allow the function to call itself recursively. However, unlike in a FunctionDeclaration, the Identifier in a FunctionExpression cannot be referenced from and does not affect the scope enclosing the FunctionExpression.

因此，如果您将代码更改为：

So if you changed your code to:

function myNinja() {
    console.log(typeof myNinja) //prints 'function'
}
var ninja = myNinja;
console.log(typeof myNinja) //prints 'function' (now we're using a declaration)

...因为它使用函数声明， myNinja 被添加到定义它的范围。 （声明也像所有声明一样被悬挂;它不像表达式那样作为逐步代码的一部分进行处理。）

...since that uses a function declaration, myNinja is added to the scope in which it's defined. (The declaration is also hoisted, like all declarations; it's not processed as part of the step-by-step code the way expressions are.)

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