Both call and apply produce the same result for String.prototype.toUppercase when passing an array

Question

I usually see that we should use an array in apply method and array of arguments in call method:

If I do the following (call method would return undefined):

var namelist = {
   f:'first name',
   l:'last name'
}
function func(a,b){
   console.log(this[a]+' '+this[b])
}

func.call(namelist,['f','l'])
//undefined undefined

func.apply(namelist,['f','l'])
//first name last name

But, look here both call and apply method works:

String.prototype.toUpperCase.call(['a','b','c']) //returns 'A,B,C'
String.prototype.toUpperCase.apply(['a','b','c']) //returns 'A,B,C'

Why the call method is working?

If I do use like this:

String.prototype.toUpperCase.call('a','b','c')
//this would return 'A' only.

Answer 1

The first argument to call and apply is the context for the function, the value of this.

But, look here both call and apply method works:

String.prototype.toUpperCase.call(['a','b','c']) //returns 'A,B,C'
String.prototype.toUpperCase.apply(['a','b','c']) //returns 'A,B,C'

Yes, that's not at all the same as your example. You're only passing one argument into each of call and apply, so they are identical. You're passing the same this with both methods. They're identical invocations.

String.prototype.toUpperCase.call('a','b','c')

//this would return 'A' only.

Yes, because you're passing only 'a' as this to toUpperCase. It would do the same thing (if it weren't an error) with apply('a', 'b', 'c'), again, because you're passing only 'a' as this to toUpperCase. The difference is that apply raises an error if you try to give it more than two arguments.

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Original answer:

That isn't how you pass arguments with call.

Use func.call(namelist, 'f', 'l').

apply takes an array of arguments. call just takes the arguments.