Javascript中的变量范围（提升） [英] Scope of variables (Hoisting) in Javascript

问题描述

  var全局变量= false; 
 
函数test（）{
 global = true; 
返回false; 
函数global（）{} 
} 
 
 console.log（global）; //说假（正如所料）
 test（）; 
 console.log（global）;如果我们假设函数在顶部被悬挂起来，那么这个函数在顶部被悬挂起来，以及var变量，让我们试试这个。
  var foo = 1; 
函数bar（）{
 return foo; 
 foo = 10; 
函数foo（）{} 
 var foo = 11; 
} 
 
 bar（）; 
 console.log（foo）; //说1（但应该是11）为什么1这次？ 
  
这是 JSBin Demo 和 function global（）{}   c>从 test（），然后运行正常。有人可以帮我理解为什么会发生这种情况吗？
      并且函数声明语句被悬挂到其封闭范围的顶部。
 
因此，函数中的函数global（）{} 会创建一个本地全球名称。
 
 
 
分配给全球函数绑定到这个本地名称。下面介绍如何使用提升来重写它，以了解编译器如何看待它： 
 
 
  function test（）{
 var global = function（）{}; //悬挂; 'global'now local 
 global = true; 
返回false; 
} 
  
 
One of my friends was taking an online quiz and he asked me this question which I could not answer.
var global = false;

function test() {
  global = true;
  return false;
  function global() {}
}

console.log(global); // says false (As expected)
test();
console.log(global); // says false (Unexpected: should be true)
If we assume that functions are hoisted at the top along with var variables, let's try this one.
var foo = 1;
function bar() {
    return foo;
    foo = 10;
    function foo() {}
    var foo = 11;
}

bar();
console.log(foo); //says 1 (But should be 11) Why 1 this time ??
Here is a JSBin Demo and JSBIN Demo2 to play with.


PS: If we remove function global() {} from test(), then it runs fine. Can somebody help me understand why is this happening ?
解决方案
var statements and function declaration statements are "hoisted" to the top of their enclosing scope.

Therefore, the function global(){} in your function creates a local global name.

Assigning to global inside your functions binds to this local name.  Here's how you can "rewrite" it using hoisting to understand how the compiler sees it:
function test() {
    var global = function() {};   // hoisted; 'global' now local
    global = true;
    return false;
}


                        
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