为什么我们在JavaScript中有Function.call？

Question

> 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).

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 Function s prototype are available in Function also.

Quoting the specification ,

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

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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.

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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.