如何在JavaScript中提升名称解析顺序？ [英] How hoisting name resolution order works in JavaScript?

问题描述

我遇到了一个有趣的测验

function bar() {
    return foo;
    foo = 10;
    function foo() {}
    var foo = '11';
}
alert(typeof bar());

我的解释是这样的（根据控制台错误:) :)：

My interpretation is like this (Which is wrong according to console :) ):

var foo; // global variable
function bar(){
    function foo(){}
    var foo;  //  Here variable foo should override foo function 
    return foo; // (according to me foo should be variable with undefined value) What is going on here, How JavaScript resolve naming order ?
    foo = 10;
    foo = "11";
}

这是我正在阅读的参考这个

Here is a reference which I am reading this

在JavaScript中，名称以四种基本方式之一进入范围：
1.Language-defined：默认情况下，所有范围都给出名称this和arguments。
2.形式参数：函数可以具有命名形式参数，其范围限定为该函数的主体。
3.函数声明：它们的形式为函数foo（）{}。
4.变量声明：这些声明采用var foo形式;

In JavaScript, a name enters a scope in one of four basic ways: 1.Language-defined: All scopes are, by default, given the names this and arguments. 2. Formal parameters: Functions can have named formal parameters, which are scoped to the body of that function. 3. Function declarations: These are of the form function foo() {}. 4. Variable declarations: These take the form var foo;

他后来引用：

要记住的最重要的特殊情况是名称解析顺序。请记住，名称有四种方式可以输入给定的范围。我在上面列出的顺序是它们被解析的顺序。通常，如果已经定义了名称，则它永远不会被同名的另一个属性覆盖。这意味着函数声明优先于变量声明。这并不意味着对该名称的赋值将不起作用，只是声明部分将被忽略。

The most important special case to keep in mind is name resolution order. Remember that there are four ways for names to enter a given scope. The order I listed them above is the order they are resolved in. In general, if a name has already been defined, it is never overridden by another property of the same name. This means that a function declaration takes priority over a variable declaration. This does not mean that an assignment to that name will not work, just that the declaration portion will be ignored.

这是令人困惑的我，有人可以参考上面的例子来简化这个吗？
要点我想知道：

Which is confusing for me, Can anyone simplify this with reference to above example ? Main point I want to know :

• 如何没有 var 的变量在一个悬挂的函数内？


• 在吊装过程中是否发生变量覆盖？

• How variables without var inside a function hoisted ?
• Does variable overriding occurs during hoisting ?

推荐答案

让我们暂时离开函数 foo 。在函数内，如果在该函数内的任何位置声明变量，则声明将移动到函数的顶部。因此，它实际上是这样评估的

Lets leave the function foo for the moment. Within a function, if a variable is declared anywhere inside that function, the declaration will be moved to the top of the function. So, it is actually evaluated like this

function bar() {
    var foo;
    return foo;
    foo = 10;
    foo = '11';
}

但是当你有一个同名的函数时，它会优先它将被评估类似于此

But when you have a function declared by the same name, it will take precedence and it will be evaluated similar to this

function bar() {
    var foo = function() {};
    return foo;
    foo = 10;
    foo = '11';
}

这就是为什么你得到函数在警告框中。

That is why you are getting function in the alert box.

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