javascript - 为什么不调用原型(prototype)中声明的函数？

Question

var p = function () {
    this.show = function () {
       alert('hello world!!!');
    }
}

p.prototype.show = function() {
    alert('haha');
}

var o  = new p();
o.show();

它提醒 “hello world!!!”，为什么？

我可以修改原型(prototype)方法吗，如果可以怎么修改？

Answer 1

那是因为您在构造函数中定义的特定函数覆盖了通过原型(prototype)继承的函数。

来自 EcmaScript specification :

Every object created by a constructor has an implicit reference (called the object’s prototype) to the value of its constructor’s “prototype” property. Furthermore, a prototype may have a non-null implicit reference to its prototype, and so on; this is called the prototype chain. When a reference is made to a property in an object, that reference is to the property of that name in the first object in the prototype chain that contains a property of that name. In other words, first the object mentioned directly is examined for such a property; if that object contains the named property, that is the property to which the reference refers; if that object does not contain the named property, the prototype for that object is examined next; and so on.

简而言之:在查找函数(或任何属性的名称)时，您从对象开始，然后在原型(prototype)链中向上移动。