Javascript继承问题

Question

我似乎无法正确覆盖类方法，使用以下代码...

function core() {
    console.log( "CORE CONSTRUCTOR CALLED" );
}

core.prototype.output = function() {
    return 'CORE';
}

function sub1() {
    console.log( "SUB 1 CONSTRUCTOR CALLED" );
    this.core();
}

sub1.prototype = core.prototype;
sub1.prototype.constructor = sub1;
sub1.prototype.core = core;

sub1.prototype.output = function() {
    return 'SUB 1';
}

function sub2() {
    console.log( "SUB 2 CONSTRUCTOR CALLED" );
    this.core();
}

sub2.prototype = core.prototype;
sub2.prototype.constructor = sub2;
sub2.prototype.core = core;

sub2.prototype.output = function() {
    return 'SUB 2';
}

var oCore = new core();
var oSub1 = new sub1();
var oSub2 = new sub2();

console.log( oCore.output() );
console.log( oSub1.output() );
console.log( oSub2.output() );

...我得到以下输出...

CORE CONSTRUCTOR CALLED
SUB 1 CONSTRUCTOR CALLED
CORE CONSTRUCTOR CALLED
SUB 2 CONSTRUCTOR CALLED
CORE CONSTRUCTOR CALLED
SUB 2
SUB 2
SUB 2

我做错了什么？

Answer 1

问题是...当您发布行时:

sub2.prototype = core.prototype;

您在 sub2 上使用相同的原型(prototype)作为core , 因此当你调用 .output()来自 任何 类，core.prototype.output 处的函数是 sub2版本，因为它是最后一个定义的。请记住，对象分配是通过引用发生的。

要复制您经常看到的对象:

sub2.prototype = new core();
sub2.prototype.core = core;

或者 - 如果您想避免调用构造函数，您可以使用 jQuery 的 $.extend(sub1.prototype, core.prototype);复制核心原型(prototype)。如果您没有 jQuery，这大致相同:

sub2.prototype = {};
for (var method in core.prototype) sub2.prototype[method] = core.prototype[method];
sub2.prototype.constructor = core;
sub2.prototype.core = core;