獲取對持有作爲構造函數的屬性的對象的引用

Question

標題確實令人困惑，我找不到更好的標題。

假設我有：

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保存到pa對象的引用。可能嗎？

Answer 1

好吧，這不是正是我想要的，但它是非常接近：

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" 

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

函數used as a constructor只有到新的實例的引用，從它的原型繼承。

要使它保持原來的A實例的引用，你就需要把B構造函數在閉包：

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 

然而，這創造了新的B構造的缺點（其自己的原型對象）爲每個單個A實例。更好的方式是像

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，這樣你只有一個原型，但不同的構造函數，我們可以混用兩種形式：

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

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