我的问题是关于维护其父对象原型(prototype)链的子对象。
在 John Resig 的 Advanced Javascript 幻灯片 (http://ejohn.org/apps/learn/#76) 中,他写道,为了维护子对象的原型(prototype)链,您必须实例化一个新的父对象。
然而,通过一些快速测试,我注意到原型(prototype)链是通过将子对象原型(prototype)设置为等于父对象原型(prototype)来维护的。
如有任何澄清,我们将不胜感激!
原始代码
function Person(){}
Person.prototype.dance = function(){};
function Ninja(){}
// Achieve similar, but non-inheritable, results
Ninja.prototype = Person.prototype;
Ninja.prototype = { dance: Person.prototype.dance };
assert( (new Ninja()) instanceof Person, "Will fail with bad prototype chain." );
// Only this maintains the prototype chain
Ninja.prototype = new Person();
var ninja = new Ninja();
assert( ninja instanceof Ninja, "ninja receives functionality from the Ninja prototype" );
assert( ninja instanceof Person, "... and the Person prototype" );
assert( ninja instanceof Object, "... and the Object prototype" );
我的修改版
function Person(){}
Person.prototype.dance = function(){console.log("Dance")};
function Ninja(){}
// Achieve similar, but non-inheritable, results
Ninja.prototype = Person.prototype;
assert( (new Ninja()) instanceof Person, "Will fail with bad prototype chain." );
var ninja = new Ninja();
assert( ninja instanceof Ninja, "ninja receives functionality from the Ninja prototype" );
assert( ninja instanceof Person, "... and the Person prototype" );
assert( ninja instanceof Object, "... and the Object prototype" );
ninja.dance();
最佳答案
在 John Resig 提供的代码中,他首先将 Ninja.prototype
设置为 Person.prototype
。然后他立即将其重置为 { dance: Person.prototype.dance }
:
// Achieve similar, but non-inheritable, results
Ninja.prototype = Person.prototype;
Ninja.prototype = { dance: Person.prototype.dance };
结果是 Ninja
构造函数创建的任何对象都将直接继承自 { dance: Person.prototype.dance }
而不是 的实例Person.prototype
.因此 (new Ninja) instanceof Person
将返回 false。在这种情况下,原型(prototype)链是:
null
^
|
| [[prototype]]
|
+------------------+
| Object.prototype |
+------------------+
^
|
| [[prototype]]
|
+------------------+
| Ninja.prototype |
+------------------+
^
|
| [[prototype]]
|
+------------------+
| new Ninja |
+------------------+
在修改后的版本中,您删除了对 Ninja.prototype
的第二个赋值,有效地将 Ninja.prototype
设置为 Person.prototype
。因此原型(prototype)链是:
null
^
|
| [[prototype]]
|
+-------------------+
| Object.prototype |
+-------------------+
^
|
| [[prototype]]
|
+-------------------+
| Ninja.prototype / |
| Person.prototype |
+-------------------+
^
|
| [[prototype]]
|
+-------------------+
| new Ninja |
+-------------------+
请注意,由于 Ninja.prototype
与 Person.prototype
相同,(new Ninja) intanceof Ninja
和 (new Ninja) instanceof Person
将返回 true
。这是因为 instanceof
operator depends on the prototype
of a constructor .
然而,在 JavaScript 中实现继承的正确方法是将 Ninja.prototype
设置为 Object.create(Person.prototype)
(或以老派方式到 new Person
),在这种情况下,原型(prototype)链将是:
null
^
|
| [[prototype]]
|
+------------------+
| Object.prototype |
+------------------+
^
|
| [[prototype]]
|
+------------------+
| Person.prototype |
+------------------+
^
|
| [[prototype]]
|
+------------------+
| Ninja.prototype |
+------------------+
^
|
| [[prototype]]
|
+------------------+
| new Ninja |
+------------------+
注意:永远记住,在 JavaScript 中,对象继承自其他对象。它们从不继承构造函数。如果您想了解 JavaScript 中真正的原型(prototype)继承,请阅读我在 why prototypal inhritance matters 上的博客文章.
关于javascript - JavaScript 中的对象继承,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/17768386/