function.prototype.method

Question

I would like to ask about the code below

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

Does that mean that "Function" and any new function will inherit functions created by method?

To make it more clear

Function.method('test', function () {return 1;});

is test now available to be called as a method on Function or any other function or not?

Answer 1

No, this inside a function refers to the object on which it was called. In this case, that should be a function, and more specifically, a constructor function. It should be used like:

function SomeObject() {}
SomeObject.method('doSomething', function() {
 alert('Something!');
});
new SomeObject().doSomething(); // Something!

• Mostafa Mahmoud: Try to define the function using the code I have attached above. After defining a function this way, I haven't been able to find it on Function's prototype or any other function. It was only available in objects created by the function and I don't know why ?!
• Ry-: @MostafaMahmoud: Ah, sorry, I misread this as Function. It’s supposed to be used on constructors. See edit.

Answer 2

JavaScript is a protypical language. When a function is called on an object and not found a search begins up the prototype chain. It will search all objects in the prototype chain until the prototype chain ends at Object, the parent of all objects.

All functions inherit from Function either directly or indirectly, meaning all functions will have your specified "method", even functions that have already been created.

---

Function.prototype.printSup = function () { console.log('sup'); }
Math.max.printSup();
String.pringSup();
'asdf'.substr.printSup()

• Ry-: The parent of all objects! Except for DOM nodes in Internet Explorer 8 and earlier, you know, because.
• Mostafa Mahmoud: Try to define the function using the code I have attached above. After defining a function this way, I haven't been able to find it on Function's prototype or any other function. It was only available in objects created by the function and I don't know why ?!
• Trent Earl: @MostafaMahmoud I dont understand, I updated my answer to include examples.