Javascript Function.prototype.call()

Question

I read some article and it said the following 2 line are doing the same thing.

fn.call(thisValue);
Function.prototype.call.call(fn, thisValue);

Answer 1

Let's start with this setup:

function fn() { console.log(this); }
var thisvalue = {fn: fn};

Now you surely understand that thisvalue.fn() is a method call, and sets the logged this value to the thisvalue object.

Next, you seem to know that fn.call(thisvalue) does exactly the same call. Alternatively, we could write (thisvalue.fn).call(thisvalue) (parentheses just for structure, could be omitted) as thisvalue.fn === fn:

thisvalue.fn(…); // is equivalent to
(thisvalue.fn).call(thisvalue, …); // or:
(fn).call(thisvalue, …);

OK, but fn.call(…) is just a method call as well - the call method of functions is called on the fn function.
It can be accessed as such because all function objects inherit this .call property from Function.prototype - it's not an own property like .fn on the thisvalue object. However, fn.call === Function.prototype.call is the same as thisvalue.fn === fn.

Now, we can rewrite that method call of .call as an explicit invocation with .call():

fn.call(thisvalue); // is equivalent to
(fn.call).call(fn, thisvalue); // or:
(Function.protoype.call).call(fn, thisvalue);

I hope you spot the pattern and can now explain why the following work as well:

Function.prototype.call.call.call(Function.prototype.call, fn, thisvalue);
var call = Function.prototype.call; call.call(call, fn, thisvalue);

Breaking this down is left as an exercise to the reader :-)