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在Javascript中,全局变量不会更改函数内部的值

[英]In Javascript, global variable is not changing value inside a function

var a = 1;

function b() {
    function a() {}; // local scope
    a = 10; // global scope
}
b();
alert(a);

It alerts 1 and not 10. I am wondering why is it so? 它警报1而不是10。我想知道为什么会这样吗?

A function name and a variable are essentially the same thing in Javascript. 函数名称和变量在Javascript中本质上是相同的。 You can declare a function via: 您可以通过以下方式声明函数:

var a = function () {};

Which is the same as function a() {} for most intents and purposes. 在大多数情况下,它与function a() {}相同。 Both create a symbol in the current scope and make that symbol's value a function. 两者都会在当前范围内创建一个符号 ,并使该符号的值成为函数。

What you're doing is you're shadowing the global a with your own local a . 你在做什么是你的阴影,全球a用自己的本地a It makes no difference whether this local a was defined via var a or function a . 此局部变量a是通过var a还是function a定义的都没有区别。

Your code is the same as this: 您的代码与此相同:

var a = 1;

function b() {
    var a = function () {}; // local scope
    a = 10;
}
b();
alert(a);

Thus, the declaration of your local scope function creates a new local variable named a that initially has a function assigned to it, but you then reassign it to the value of 10 . 因此,您的局部作用域函数的声明将创建一个名为a的新局部变量,该变量最初具有分配给它的函数,但随后将其重新分配为值10 The higher scoped a is not touched by this inner assignment. 此内部分配未触及较高范围的a

If the outer a definition is in the global scope, then you could assign to it with: 如果外a定义是在全球范围内,那么你可以分配给它:

 window.a = 10;

If it is not in the global scope, then it has been "hidden" by the inner definition of a and there is no way to reach the outer a directly from the inner scope. 如果它不在全局范围内,则它已被a的内部定义“隐藏”,并且无法直接从内部范围到达外部a

JavaScript is different from other languages: JavaScript与其他语言不同:

JavaScript® (often shortened to JS) is a lightweight, interpreted, object-oriented language with first-class functions JavaScript®(通常缩写为JS)是一种轻量级,解释性,面向对象的语言,具有一流的功能

What is first-class ? 什么是first-class

allows functions to be passed around just like any other value. 允许像其他任何值一样传递函数。

So as jfriend00 pointed out it converts the function into a variable locally in the function thus not changing the global variable. 因此,正如jfriend00指出的那样,它将函数转换为函数中的局部变量,因此不会更改全局变量。

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