是真的，JavaScript中的每个函数都是一个闭包吗？ [英] Is it true that every function in JavaScript is a closure?

问题描述

我理解JavaScript中的每个函数都是一个第一类对象，它有一个内部属性[[scope]]，它托管函数的自由变量的绑定记录。但是，有两种特殊情况。

1. Function构造函数创建的函数是否也是一个闭包？ Function构造函数创建的函数对象是特殊的，因为它的[[scope]]可能不会引用其外部函数的词法环境，而只引用全局上下文。例如，

     var a = 1; 
    var fn =（function outer（）{
    var a = 2; 
    var inner = new Function（'alert（a）;'）; 
    return inner; 
   }）（）; 
    fn（）; // will alert 1，not 2. 
     

   这是不直观的。这也被称为闭包？




2. 如果内部函数没有任何自由变量，我们可以说在创建内部函数时形成闭包？例如，

     //这是一个仅用于学术研究的无用的情况
    var fn =（function outer {
    var localVar1 = 1，
    localVar2 = 2; 
    return function（）{}; 
   }）（）; 
     

   在这种情况下，fn是指作为内部函数创建的空函数对象。它没有自由变量。在这种情况下，我们可以说形成了关闭吗？

解决方案

由Function构造函数创建的函数是否也是一个闭包？

这可能是不直观的，因为所有其他JavaScript闭包在它们的词汇范围上，但它仍然匹配我们的定义封闭件。在您的示例中， a 是一个自由变量，并在其他作用域中解析为 a $ c> inner / fn 函数被调用。

如果内部函数没有任何自由变量，我们仍然可以将其称为闭包？

问。 一些说是，其他称为无趣的关闭，我个人说，因为他们不引用一个外部范围。 p>

I understand that every function in JavaScript is a first-class object and it has an internal property [[scope]] which hosts the binding records of the function's free variables. However, there are two special cases.

1. Is the function created by Function constructor also a closure? The function object created by Function constructor is special, because its [[scope]] may not refer to the lexical environments of its outer functions, but only the global context. For example,

   var a = 1; 
   var fn = (function outer() {
       var a = 2; 
       var inner = new Function('alert(a); ');
       return inner;
   })();
   fn(); // will alert 1, not 2.
   

   This is unintuitive. Is this also called closure?

2. If an inner function doesn't have any free variables, can we say a closure is formed when the inner function is created? For example,

   // This is a useless case only for academic study
   var fn = (function outer() {
       var localVar1 = 1,
           localVar2 = 2;
       return function() {};
   })();
   

   In this case, fn refers to an empty function object which was created as an inner function. It has no free variables. In this case can we say a closure is formed?

解决方案

Is the function created by Function constructor also a closure?

Yes, it closes over the global scope. That might be unintuitive because all other JavaScript closures close over their lexical scope, but it still matches our definition of a closure. In your example, a is a free variable, and resolves to the a in an other scope when the inner/fn function is called somewhere.

If an inner function doesn't have any free variables, can we still call it a closure?

Depends on whom you ask. Some say Yes, others call them "uninteresting closures", personally I say No because they don't reference an outer scope.

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