JavaScript 中的 `new` 有什么作用呢? [英] What does `new` in JavaScript do, anyway?

问题描述

我对构造函数在 Javascrpt 中的工作方式感到非常困惑；尽管使用该语言已有数年(主要就好像它就像 LISP 的半命令式版本)，但我想更多地了解对象应该如何在其中工作.

鉴于此代码:

function Foo(x) {返回 {酒吧:函数(){返回x；}};}

调用 myFoo = Foo(5) 和 myFoo = new Foo(5) 有什么区别?或者，换句话说，Javascript 中的构造函数究竟做了什么do?

解决方案

调用 myFoo = Foo(5) 和 myFoo = new Foo(5) 有什么区别?

该代码没有区别，因为它返回一个对象，并且规范说:

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• 让 result 是调用 F 的 [[Call]] 内部属性的结果，提供 obj 作为 this 值并提供作为参数传入 [[Construct]] 的参数列表.
• 如果 Type(result) 是 Object，则返回 result.

由于该函数返回的结果是一个对象，因此使用其结果.如果它没有返回一个对象，或者如果它检查了this，你会注意到一个不同，例如如果你把它改写为:

function Foo(x) {if (!(this instanceof Foo)) { return new Foo(x);}this.bar = function() { 返回 x;};}//现在 instanceof 起作用了.alert((new Foo) instanceof Foo);

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JavaScript 中的 new 到底做了什么?

new 操作符导致函数被调用，this 绑定到一个新创建的 Object，其原型是该函数的 prototype 属性.

对于用户定义的函数，

new f(a, b, c)

相当于

//使用 f 的原型创建一个新实例.var newInstance = Object.create(f.prototype), result;//调用函数结果 = f.call(newInstance, a, b, c),//如果结果是非空对象，则使用它，否则使用新实例.结果&&结果类型 === '对象' ?结果:新实例

注意，语言规范实际上定义了两个操作的函数，[[Call]] 和 [[Construct]]，所以在某些极端情况下 new 的行为很奇怪.

例如绑定和内置函数:

var g = f.call.bind(f);

应该定义一个函数，调用时只调用f，所以g在各方面都应该和f相同，但是

new g()

生产

TypeError: function call() { [native code] } 不是构造函数

因为内置函数 Function.prototype.call 支持 [[Call]] 但不支持 [[Construct]].

Function.prototype.bind 在 new 和常规调用方面的行为也不同.this 值在调用时始终是绑定的 thisValue，但在使用 new 时是新构造的实例.

I am very confused about how constructors work in Javascrpt; despite using the language for several years (mostly as if it were like a semi-imperative version of LISP) I would like to know more about how objects are supposed to work in it.

Given this code:

function Foo(x) {
    return {
        bar: function() { return x; }
    };
}

What is the difference between calling myFoo = Foo(5) and myFoo = new Foo(5)? Or, in other words, what exactly does a constructor in Javascript do?

解决方案

What is the difference between calling myFoo = Foo(5) and myFoo = new Foo(5)?

There's no difference for that code, because it returns an object, and the spec says:

• Let result be the result of calling the [[Call]] internal property of F, providing obj as the this value and providing the argument list passed into [[Construct]] as args.
• If Type(result) is Object then return result.

Since that function returns a result that is an Object, its result is used. You would notice a difference if it did not return an object, or if it checked this, for example if you rewrote it as:

function Foo(x) {
  if (!(this instanceof Foo)) { return new Foo(x); }
  this.bar = function() { return x; };
}
// Now instanceof works.
alert((new Foo) instanceof Foo);

What does new in JavaScript do, anyway?

The new operator causes the function to be called with this bound to a newly created Object whose prototype is that function's prototype property.

For user-defined functions,

new f(a, b, c)

is equivalent to

// Create a new instance using f's prototype.
var newInstance = Object.create(f.prototype), result;

// Call the function
result = f.call(newInstance, a, b, c),

// If the result is a non-null object, use it, otherwise use the new instance.
result && typeof result === 'object' ? result : newInstance

Note, that the language specification actually defines functions with two operations, [[Call]] and [[Construct]], so there are some corner cases where new behaves oddly.

For example, bound and built-in functions:

var g = f.call.bind(f);

should define a function that when called, just calls f, so g should be the same as f in all respects, but

new g()

produces

TypeError: function call() { [native code] } is not a constructor

because the builtin function Function.prototype.call supports [[Call]] but not [[Construct]].

Function.prototype.bind also behaves differently around new and regular calls. The this value is always the bound thisValue when called, but is a newly constructed instance when you use new.

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