原型继承的构造函数赋值

问题描述

在进行原型继承时，会要求将子构造函数引用回自身，

When making the prototypal inheritance, it's asked to refer the child's constructor back to itself below,

A = function() {}
B = function() {}
B.prototype = new A;
B.prototype.constructor = B;

如果不是会产生什么不利影响？

What would take adverse effect if not?

编辑

• 作为 @GGG & @CMS 已经解释过，构造函数对齐对于通过 new Child（...）效果c $ c>，但是后来正确反映子对象的构造函数所需的是
。


• @GGG 也有建议作为原型链扩展
  的防御性替代方案。而 Child.prototype = new Parent（...）
  包括父级的子属性， Child.prototype =
Object .create（Parent.prototype）没有。

• As @GGG & @CMS have explained, the constructor alignment takes noeffect on creating the child object by new Child(...), but isnecessary to correctly reflect the child object's constructor later.
• @GGG has also suggested a defensive alternative to extendthe prototype chain. While Child.prototype = new Parent(...)includes parent's properties to child, Child.prototype =Object.create(Parent.prototype) doesn't.

推荐答案

首先，请不要这样做：

B.prototype = new A;

改为执行此操作（ 旧浏览器的Object.create ：

B.prototype = Object.create(A.prototype);

对于构造函数，如果没有任何内容会破坏你不这样做，但如果你不这样做：

As for constructor, nothing will break if you don't do this, but if you don't:

A = function() {};
var a = new A();
console.log(a.constructor); // A

B = function() {};
var b = new B();
console.log(b.constructor); // A (!)

...设置构造函数原型的属性回到实际的构造函数允许你这样做：

...setting the constructor property of the prototype back to the actual constructor function allows you to do this:

B.prototype.constructor = B;
var b = new B();
console.log(b.constructor); // B

这篇关于原型继承的构造函数赋值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！