为什么我不能调用function.apply？

Question

This is related to this question: Is it possible to spread the input array into arguments?

I'm assuming that given this line of code:

Promise.all(array).then(foo)

Promise.all uses Function.call , to invoke foo ,

foo.call(foo, arrayValues)

I would like to modify foo into a foo.apply function such that calling it with an array of values splits it into regular arguments.

Here is my train of thought....

Assuming I have this function

function test(a,b,c){
    console.log(a,b,c)
}

I can call this function using both call and apply

test.call(null,1,2,3)
>> 1 2 3
test.apply(null,[1,2,3])
>> 1 2 3

So far so good, this also works...

test.call.apply(test,[null,1,2,3])
>> 1 2 3

However I cannot get this to work

test.apply.call(test,[null,1,2,3])
>> undefined undefined undefined

What is happening here??

Answer 1

I got it to work

test.apply.call(test,null,[1,2,3])
>> 1 2 3

Answer 2

test.apply.call(test,[null,1,2,3])

equals to

test.apply([null,1,2,3])

equals to

test()

So you got undefined as output.

---

test.apply.call(test,null,[1,2,3])

equals to

test.apply(null,[1,2,3])

equals to

test(1,2,3)

This is right.