TypeError：调用Function.prototype.method（）时未定义this.prototype

Question

I am reading the book "Javascript: The good parts".
Now I am reading chapter about Augmenting Types:

Function.prototype.method = function (name, func) {
   this.prototype[name] = func;
   return this;
};

UPDATE:
Why following code does not work?

js> Function.prototype.method("test", function(){print("TEST")});
typein:2: TypeError: this.prototype is undefined

But following code works without problems:

js> Function.method("test", function(){print("TEST")});
function Function() {[native code]}

Why this code works?

js> var obj = {"status" : "ST"};
js> typeof obj;
"object"
js> obj.method = function(){print(this.status)};
(function () {print(this.status);})
js> obj.method();
ST

"obj" is object.
But I can call method "method" on it.
What is the difference between Function.prototype.method and obj.method?

Answer 1

Because:

Function instanceof Function           // <--- is true
Function.prototype instanceof Function // <-- is false

• Function.prototype is an Object, and does not inherit anything from the Function contructor.
• Function is a constructor, but also a function, so it inherits methods from Function.prototype .
  
  
• When calling Function.method , you're calling the method method of an instance of Function . So, this points to the created instance of Function .
• When calling Function.prototype.method , you're invoking an ordinary method of an object. this points to Function.prototype .

To clarify, here's an example:

Function.method()                // is equivalent to
(function Function(){}).method()
(new Function).method()          // Because Function is also a function

Function.prototype.method // No instance, just a plain function call

Answer 2

this refers to Function.prototype because you called .method on that. So, you're using Function.prototype.prototype which does not exist.

Either use Function.method(...) or this[name] = ... to eliminate one of the .prototype s.

Answer 3

prototype is only used when you are declaring the function, but not when you are calling it. Prototype makes the function a member of the object that gets created with every instance of that object.