为什么调用 Function.apply.bind(fn, null) 调用的是 `fn.apply`，而不是 `Function.apply`？

Question

Suppose we have this code:

let fn1 = Function.apply.bind(Math.max, null);
fn1([1, 10, 5]); // returns 10

I know that it's a spread function in ES6 we would instead write:

Math.max(...[1, 10, 5]);

But I can not understand how Function.apply.bind does its magic...

This is how I think code works, please correct me where I am wrong.

1. This statement let fn1 = Function.apply.bind(Math.max, null); creates a new function Function.apply(null) with this set to Math.max and assigns it to fn1 variable.
2. This statement fn1([1, 10, 5]); invokes our newly created function in this manner: Function.apply(null, [1, 10, 5]) with this inside this function set to Math.max . But this call is invalid... if we copy/paste that line in JS console - we will get Uncaught SyntaxError: Unexpected number

So, my question is - how Function.apply knows that it should use this equal to Math.max and "rotate"/modify its call to Math.max.apply ?

Updated

I just understood how all that stuff works! First, let's rewrite a code to its "real" form:

let fn1 = this.Function.apply.bind(Math.max, null);
this.fn1([1, 10, 5]); // returns 10

Magic? this is implicitly added to "global" methods/objects. Now let's rewrite those 2 paragraphs of text to show you how magic works :)

1. This statement let fn1 = Function.apply.bind(Math.max, null); creates a new function newFunction.apply(null, ...args) , where newFunction == Math.max , so it's the same as a function Math.max.apply(null, ...args) and assigns it to fn1 variable.
2. This statement fn1([1, 10, 5]); calls newFunction.apply(null, [1, 10, 5])

Magic revealed :)

Duplicate update

This question is not a duplicate of another question. It's different because it's more concrete, here I am asking specifically "Why call to Function.apply.bind(fn, null) calls fn.apply , and not Function.apply ?" and not "How apply.bind works?". It's like saying that a question "Why bicycle gears use kinesthetic force of pedals and not directly generate force by acceleration?" is the same as a question "How does a bicycle work?"

Answer 1

Let's split it into parts, first of all, bind() :

The bind() method creates a new function that, when called, has its this keyword set to the provided value, with a given sequence of arguments preceding any provided when the new function is called.

So you are bind ing the apply method from Function global object to Math.max .

The function fn1 returned by bind() is therefore equivalent to this.apply() , where this really is Math.max , exactly as you "asked" to the bind() method.

Answer 2

This line:

let fn1 = Function.apply.bind(Math.max, null);

...is really tricky. :-) It calls bind on Function.apply , creating a function that will call apply with Math.max as the this value and null as the first argument (the thisArg that apply should use when calling Math.max ; Math.max doesn't care about this , so null is fine).

When you call fn1 , it calls Function.apply with this set to Math.max , and null as the first argument, followed by the array you pass into fn1 . apply calls Math.max with null as this and the entries from the array as discrete arguments.