Why if F - simple function: F.prototype!== F.__proto__ but Function.prototype === Function.__proto__?

Question

Why if F - simple function:

F.prototype !== F.__proto__

but

Function.prototype === Function.__proto__ 

?

Answer 1

F.prototype !== F.__proto__

Suppose you're designing an API for all functions. So you define that every function should have the method call. You create an object with such method:

var fproto = {call: ()=>{}};

Then for all functions to share this functionality, you have to add it to .prototype property of a Function constructor, so that all instances of a Function inherit it. So you do the following:

Function.prototype = fproto.

Now, when you create a function F, it will have have its .__proto__ set to fproto:

const F = new Function();
F.call(); // works because of lookup in prototype chain through `__proto__` property
F.__proto__ === Function.prototype; // true

Now you decide that all instances created using F constructor, should have a method custom, so you create an object iproto with the property and set it as a prototype for all instances of F using prototype property:

const iproto = {custom: ()=>{}};
F.prototype = iproto;

const myobj = new F();
myobj.custom(); // works

So now it should be clear that F.__proto__ and F.prototype are not the same object. And this is essentially what happens under the hood when you declare a function:

const F = function() {};

// F.__proto__ is set to Function.prototype to inherit `call` and other methods
F.__proto__ === Function.prototype

// F.prototype is set to a new object `{constructor: F}` and so:
F.prototype !== Function.prototype

Function.prototype === Function.__proto__

Is an exceptional case because Function constructor should have all methods available for function instances, and hence Function.__proto__, but all share these methods with function instances, hence Function.prototype.