在这段代码中：
var Fruit = function() {}
Fruit.prototype = {
color: function () {
console.log('Fruit color...')
}
}
var Apple = function () {}
Apple.prototype = new Fruit()
Apple.prototype.constructor = Apple
var a = new Apple()
Apple.prototype = null // the question!!!
a.color()
解决方法
在创建实例a后,您将更改Apple.prototype引用.
此处更改引用不会更改现有实例的引用.
你也会发现
var a = new Apple();
Apple.prototype = {}; // some other object
a instanceof Apple; // false
即因为我们已经改变了Apple的继承链,因此不再被视为Apple.
如果您尝试进行Foo检查实例,设置Foo.prototype = null将导致TypeError
更改Object的属性不会更改对该Object的引用.例如
var foo = {},bar = foo;
foo.hello = 'world';
foo === bar; // true
更改对象本身会更改引用
foo = {hello: 'world'};
foo === bar; // false
或者以更接近于如何从实例引用原型的方式编写,
var Foo = {},// pseudo constructor
bar = {},baz = {};
var fizz = {}; // fizz will be our pseudo instance
Foo.bar = bar; // pseudo prototype
fizz.inherit = foo.bar; // pseudo inheritance
Foo.bar = baz; // pseudo new prototype
fizz.inherit === foo.bar; // false,instance inheritance points elsewhere
设置继承链的当前最佳实践不是使用new,而是使用Object.create
Apple.prototype = Object.create(Fruit.prototype);
如果你需要在Apple实例上调用Fruit构造函数,你会这样做
function Apple() {
// this instanceof Apple
Fruit.apply(this);
// ...
}
