JS寄生组合继承的疑问

Question

以下为JS寄生组合继承的实现方式         function Super(b){             this.b = b;             this.fun = function(){}         }         Super.prototype.c = function(){console.log(1111)}         function Foo(a,b){             this.a = a;             Super.call(this,b);         }         var f = new Function();         f.prototype = Super.prototype;         Foo.prototype = new f();         var foo1 = new Foo(1,2);为什么不直接用以下方式，更简洁而且能实现同样的效果        function Super(b){             this.b = b;             this.fun = function(){}         }         Super.prototype.c = function(){console.log(1111)}         function Foo(a,b){             this.a = a;             Super.call(this,b);         }         Foo.prototype = Super.prototype;         var foo1 = new Foo(1,2);

Answer 1

你是不是打错了
Foo.prototype = Foo.prototype; Foo.prototype = Super.prototype;

Foo.prototype = Super.prototype;
Foo.prototype.c = function(){console.log(1111)}
和
Foo.prototype = new f();

还是不一样吧