> Function.call == Function.prototype.call
true
> Function.prototype == Function
false

Why do Function.prototype.* methods exist as Function.*? It seems inconsistent.

This isn't the case with any other primary type (Array.slice doesn't exist but Array.prototype.slice does).

> Function.call == Function.prototype.call
true
> Function.prototype == Function
false

Why do Function.prototype.* methods exist as Function.*? It seems inconsistent.

This isn't the case with any other primary type (Array.slice doesn't exist but Array.prototype.slice does).

• javascript

1 Answer 1

Because Function itself is the prototype of Function

console.log(Function instanceof Function);
console.log(Object.getPrototypeOf(Function) === Function.prototype);

So, all the functions in the Functions prototype are available in Function also.

Quoting the specification,

The Function prototype object is itself a Function object (its [[Class]] is "Function")

Another way to confirm this would be,

console.log(Function.call === Function.prototype.call);

it means that the Function.call object and Function.prototype.call object are the same.

console.log(Function.hasOwnProperty('call'));

it means that the Function object itself doesn't have call property.

console.log(Function.prototype.hasOwnProperty('call'));

it means that Function.prototype object has the call property.

Array.slice doesn't exist but Array.prototype.slice do

Because Array function's prototype is Function object, not the Array object.

console.log(Object.getPrototypeOf(Array) === Function.prototype);

That is why we get call, apply, bind etc on the Array function. It Array object had been the prototype of Array, then slice would have been available on the Array object also.