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