原型继承。 obj-> C-> B-> A，但obj.constructor是A.为什么？

Question

var print = function(text){ 
    document.write(text); 
    document.write("</br>"); 
} 

var A = function(){ 
} 
A.prototype.name="A"; 

var B = function(){ 
} 
B.prototype = new A(); 
B.prototype.name="B"; 

var C = function(){ 
} 
C.prototype = new B(); 
C.prototype.name="C"; 

obj = new C(); 
print(obj.name); 
print(obj.constructor.prototype.name); 
print(obj.constructor == A); 

该代码给出了一个输出：

C 
A 
true 

为什么obj.constructor这里是A和不是C？

Answer 1

如this code sample看到的，你必须使用继承时，或你的构造被覆盖，当你调用new A()或new B()手动重置.constructor属性：

B.prototype = new A(); 
B.prototype.constructor = B; // need this line to fix constructor with inheritance 

这里是工作示例：http://jsfiddle.net/93Msp/。

希望这会有所帮助！

Answer 2

我认为这篇文章解释了原型继承有关构造函数的一个很好的协议

http://phrogz.net/js/classes/OOPinJS2.html

Answer 3

为了使画面清晰：

在链中唯一

obj->"new B()"->"new A()" // where obj is the same as "new C()" 

"new A()"对象有财产constructor。所有其他对象从原型链获得constructor属性。

在代码：

var A = function(){ 
} 
A.prototype.name="A"; 
// A had not create "constructor" property for "new A()" 
// so, looking at the prototype 
// According to ECMAScript spec 13.2.10 
// A.prototype.constructor == A 
// thus 
// "new A()".constructor == A 

var B = function(){ 
} 
B.prototype = new A(); 
B.prototype.name="B"; 
// B had not create "constructor" property for "new B()" 
// looking at the prototype 
// B.prototype is "new A()" 
// as shown above 
// "new A()".constructor == A 
// thus 
// B.prototype.constructor == A 
// and at the end 
// "new B()".constructor == A 

var C = function(){ 
} 
C.prototype = new B(); 
C.prototype.name="C"; 
// C had not create "constructor" property for "new C()"/"obj" 
// looking at the prototype 
// C.prototype is "new B()" 
// looking up to "new B()".prototype 
// "new B()".prototype is "new A()" as shown above 
// "new A()".constructor == A 
// and finally 
// C.prototype.constructor == A 

obj = new C(); 
print(obj.name); 
print(obj.constructor.prototype.name); 
print(obj.constructor == A); 

因此，作为写mellamokb我们应该覆盖（如果更精确的创建，）constructor财产。