CODEWITHSUNDEEP
javascriptprototype
doubleOrt
doubleOrt

Reputation: 2507

Why does Function.__proto__ return something different than other prototypes?

If you try to print out the __proto__ property of a regular object, say {}.__proto__ or foo.__proto__, you get [object object]; this [object object] is the prototype of another object. However, if you try to print out the __proto__ property of any function, it gives you function () { [native code] }. Why is that ? Shouldn't it return the prototype of the object that Function inherits from, which should be an object instead of another function (like function(){native code]}) ? How is it possible that the __proto__ of a function is another function and not an object, like other prototypes ?

*i know that Object.getPrototypeOf(obj) should be used instead of proto, but i used that instead because it is shorter.

If someone explained the above to me, i would be grateful as well. And if you have any confusions regarding my question, please ask in the comments instead of downvoting.

Upvotes: 3

Views: 991

Answers (3)

Shawn
Shawn

Reputation: 57

All the 'uppercase' constructor functions' _ proto _ will refer to Function. prototype or Function._ proto _(they refer to the same object) which is function () { [native code] } you get.

Upvotes: 0

fgb
fgb

Reputation: 18569

In JavaScript, calling a constructor function with var o = new f() creates a new object, and sets o.__proto__ to f.Prototype. Same as:

new Array().__proto__ == Array.prototype // true
new Number().__proto__ == Number.prototype // true

or:

[].__proto__ == Array.prototype // true
(5).__proto__ == Number.prototype // true

Function is a constructor function, so creating a function with new Function() sets its prototype to Function.prototype:

new Function().__proto__ === Function.prototype // true

or:

(function() {}).__proto__ == Function.prototype // true

All functions have this same prototype, including constructor functions. So:

Function.__proto__ == Function.prototype // true
Array.__proto__ == Function.prototype // true
Number.__proto__ == Function.prototype // true

Function.prototype defines the default behavior of functions, that all functions inherit from, including the ability to call it, so it acts as a function itself:

Function.prototype() // undefined
(new Function())() // undefined

Upvotes: 0

Jonas Wilms
Jonas Wilms

Reputation: 138277

Cause

Function.prototype.toString()

will return 

"function (){ [native code] }"

The Function.prototype is an object, but as you print it, its typecasted to a string, and as the prototype implements the behaviour of functions:

function(){}.toString()

it will print being a function even if it isnt.

function(){}
 .__proto__ // Function.prototype
 .__proto__ // Object.prototype
 .toString() // "[Object object]

Maybe more imaginable:

class Human {
  toString(){
    return "Im a Human";
  }
}

console.log( 
(new Human).__proto__
);
//the Human.prototype tells you that it is a Human, however its just a class.

Upvotes: 3

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