TypeError: this.prototype is undefined when calling Function.prototype.method()

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?

pimvdb · Accepted Answer · 2012-02-08 18:41:10Z

5

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 .prototypes.

answered Feb 8, 2012 at 18:41

pimvdb

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1 Comment

Volodymyr Bezuglyy Over a year ago

Thank you. Now I understand. I did not pay attention on following line this.prototype[name] = func;

Rob W · Accepted Answer · 2012-02-08 21:09:51Z

5

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

edited Feb 8, 2012 at 21:09

answered Feb 8, 2012 at 18:37

Rob W

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3 Comments

pimvdb Over a year ago

"When calling Function.prototype.method, you're invoking an ordinary function. this points to undefined (strict mode) or window." - this is not correct. It points to the object that you're calling it on; in this case, Function.prototype.

Volodymyr Bezuglyy Over a year ago

@pimvdb So what's the difference between object Function.prototype and object obj from my question?

Rob W Over a year ago

@VladimirBezugliy When calling Function.prototype.method(), this points to an object, which doesn't have a prototype property. Type this in your REPL: ({}).prototype. The return value will be undefined.

Kory Hodgson · Accepted Answer · 2012-02-08 18:39:27Z

0

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.

answered Feb 8, 2012 at 18:39

Kory Hodgson

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