Question

不知何故，JS继承并没有在我脑海中浮现。我现在尝试了很多东西，由于某种原因，我无法向继承的对象添加方法。

请参阅我的定义B.setC的尝试，然后以某种方式无法使用。任何提示？

欢呼声汤姆

A = function( value){
        this.aValue = value;
}


B = function(value){
    A.call(this, value); 
    this.bValue=value+value;
}

B.prototype = Object.create(A.prototype);//subclass extends superclass


B.setC = function(value){
    this.c = value+value+value;
    console.log("execution");
}

B.prototype.setD = function(value){
    this.D = value+value+value+value;
    console.log("execution");
}





 x = new B("ConstructorParam");
x.setC("methodParam");// gives that setC is not a function
x.setD("methodParam");// works but I added setD to A not B

Answer 1

混淆似乎来自这种误解;引用你的评论：

实际上这不是拼写错误。我想将setC添加到B而不是A，如果我的理解是正确的，并且我在Chrome中看到的调试是，如果我使用B.prototype.setB然后我将setB添加到B又名A的原型而不是B。或者我错过了吗？

这个评论中也显示了这个问题：

x.setD("methodParam");// works but I added setD to A not B

B.prototype不是A。 B.prototype是一个以A.prototype为原型的对象。您已将setD正确添加到B.prototype;它不以任何方式添加到A或A.prototype。

如果您希望按照展示方式使用setC和setD，则可以将它们放在B.prototype上（因此可以在B的实例上访问它们）在A.prototype上（因此可以在A或B的实例上访问它们。

如果我们将B.setC =更改为B.prototype.setC =，请将d小写设置为匹配c，添加一些缺少的var，并在创建时使用较短的值并调用方法，我们得到这个：

var A = function( value){
    this.aValue = value;
};

var B = function(value){
    A.call(this, value); 
    this.bValue = value + value;
};

B.prototype = Object.create(A.prototype);//subclass extends superclass

B.setC = function(value){
    this.c = value + value + value;
    console.log("execution");
};

B.prototype.setD = function(value){
    this.d = value + value + value + value;
    console.log("execution");
};

var x = new B("b");
x.setC("c");
x.setD("d");

在最后一行代码之后，这就是我们在内存中的内容（减去一堆不必要的细节）：

         +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
         |                                                               |
         \ +−−−−−−−−−−−−−−−+                                             |
A−−−−−−−−−>|   function    |                                             |
           +−−−−−−−−−−−−−−−+                             +−−−−−−−−−−−−−+ |
           | prototype     |−−−−−−−−−−−−−−−−−−−−−−−−−−−−>|   object    | |    
           +−−−−−−−−−−−−−−−+                           / +−−−−−−−−−−−−−+ |    
                                                       | | constructor |−+
                                                       | +−−−−−−−−−−−−−+
         +−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+ |
         |                                           | |
         \ +−−−−−−−−−−−−−−−+                         | |
B−−−−−−−−−>|   function    |                         | |
           +−−−−−−−−−−−−−−−+    +−−−−−−−−−−−−−−−−−−+ | |
           | prototype     |−−−>|      object      | | |
           +−−−−−−−−−−−−−−−+  / +−−−−−−−−−−−−−−−−−−+ | |
                              | | constructor      |−+ |
                              | | setC: (function) |   |
                              | | setD: (function) |   |
                              | | [[Prototype]]    |−−−+
           +−−−−−−−−−−−−−−−+  | +−−−−−−−−−−−−−−−−−−+
x−−−−−−−−−>|    object     |  |
           +−−−−−−−−−−−−−−−+  |
           | aValue: "a"   |  |
           | bValue: "aa"  |  |
           | c: "ccc"      |  |
           | d: "dddd"     |  |
           | [[Prototype]] |−−+
           +−−−−−−−−−−−−−−−+

上面的

[[Prototype]]是规范用于包含对其原型对象的引用的对象的“内部槽”的名称。相比之下，prototype，函数上的属性（例如，A.prototype），只是函数的正常属性，指向new将用作[[Prototype]]的对象如果将该函数与new一起使用，则会创建新对象。

为了完整起见，这是在ES2015 +：

class A {
    constructor(value) {
        this.aValue = value;
    }
}

class B extends A {
    constructor(value) {
        super(value); 
        this.bValue = value + value;
    }

    setC(value) {
        this.c = value + value + value;
        console.log("execution");
    }

    setD(value) {
        this.d = value + value + value + value;
        console.log("execution");
    }
}

let x = new B("b");
x.setC("c");
x.setD("d");

Javascript调用函数中的继承

1 个答案: