为什么函数的原型是函数，为什么对象是从那里继承的？

Question

Function.prototype // function Empty() {}

这有什么意义？例如，如果我们使用Number对象，我们可以看到他的原型(Number.__proto__)是Function.prototype，它包括apply和call之类的方法。如果Number的原型是一个空函数而不是与所有其他原型一样的常规原型对象，我如何使用Number.apply(..)？(数字原型、字符串原型、任何其他自定义原型都是对象。甚至Object.prototype也是一个对象)。

在那之后，Object.__proto__ == Function.prototype又有什么意义呢？对象应该是最高的对象，当Function.prototype从.继承时，它是如何从Function继承的。Object.prototype！

Object instanceof Function // true
Function instanceof Object // of course true
Function instanceof Function // true

Answer 1

米克洛斯是对的，但更简单地说：

Object.__proto__ == Function意味着Object本身就是一个函数，因为它是构造函数。这并不意味着从Object继承的对象将继承Function。对象继承构造函数的.prototype，而不是.__proto__

换句话说

function Car (){}
inst = new Car ();
// inst inherits from Car.prototype
// inst.__proto__ == Car.prototype;
// Car inherits from Function.prototype because it is a function
// Car.__proto__ == Function.prototype;

但这并不意味着inst继承自Function.prototype，您不能在其上调用apply和call。

// This means that Everything that inherits from function will
console.log(`Function.prototype`) === function Empty() {}

又一次扭转

// This means that the constructor function (Object)
// inherits from `Function.prototype` That is, you can use call and apply,
// And at a lower language level, you can use () and new on it.
Object instanceof Function // true

// It doesn't mean that instances created from Object inherit 
// from Function.prototype (can't use call/apply)
(new Object()) instanceOf Function ? // false
(new Object()).apply === undefined ? // true

// This means that functions themselves are objects, everything is an object
// They have properties like hasOwnProperty and isPrototypeOf
// Not that everything that inherits from Object.prototype will also inherit
// From Function.prototype
Function instanceof Object // of course true