Question

标题真的令人困惑，我无法找到更好的标题。

假设我有：

var A = function (){
    this.pa = { x: 1 };
};

A.prototype.B = function (){
    this.pb = /* a reference to { x: 1 } */;
};

var a = new A ();
var b = new a.B ();
console.log (b.pb.x); //should print 1
a.pa.x = 2;
console.log (b.pb.x); //should print 2

我想在pb中保存对pa对象的引用。有可能吗？

Answer 1

函数used as a constructor只有对新实例的引用，继承自其原型。

要使其保持对原始A实例的引用，您需要将B构造函数放在闭包中：

function A() {
    var that = this;
    this.pa = { x: 1 };

    this.B = function() {
        this.pb = that.pa;
    };
};

var a = new A ();
var b = new a.B ();
console.log (b.pb.x); // does print 1
a.pa.x = 2;
console.log (b.pb.x); // does print 2

但是，这样做的缺点是为每个B实例创建一个新的A构造函数（带有自己的原型对象）。更好的是像

function A() {
    this.pa = { x: 1 };
}
A.B = function() {
    this.pb = null;
};
A.prototype.makeB = function() {
    var b = new A.B();
    b.pb = this.pa;
    return b;
};
// you can modify the common A.B.prototype as well

var a = new A ();
var b = a.makeB();
console.log (b.pb.x); // does print 1
a.pa.x = 2;
console.log (b.pb.x); // does print 2

然而，我们可以混合使用这两种方法，这样你只有一个原型但不同的构造函数：

function A() {
    var that = this;
    this.pa = { x: 1 };

    this.B = function() {
        this.pb = that.pa;
    };
    this.B.prototype = A.Bproto;
}
A.Bproto = {
    …
};

Answer 2

var A = function (){
    this.pa = { x: 1 };
};

A.prototype.B = function (a){

     this.pb = a.pa;
};
var a = new A ();

var b = new  a.B(a);
console.log(b.pb.x); //should print 1
a.pa.x = 2;
console.log(b.pb.x);

Answer 3

嗯，这不是我想要的，但它非常接近：

var A = function (pa){
    this.pa = pa;
};

A.prototype.B = function (a){
    if (this instanceof A.prototype.B){
        if (!a) throw "error";
        this.pb = a.pa;
        return;
    }
    return new A.prototype.B (this);
};

var a = new A ({ x: 1 });
var b = a.B ();
console.log (b.pb.x); //1
a.pa.x = 2;
console.log (b.pb.x); //2

new a.B () //throws "error"

获取对包含属性的对象的引用，该属性是构造函数

3 个答案: