为什么会这样呢？

Question

I recently watched a video where the teacher demonstrated that the javascript code

alert.call.apply(function(a) {return a}, [1,2])

results always in "2". So here is my question: Why??

I could not find any explination why this result is returned.

Answer 1

This code is just too clever! But wait, the following code will produce the same result:

console.log.call.apply(function(a) {return a}, [1,2])

Here is what's happening: apply() 's first parameter is a ' this ', the second is a list of argument. It then calls the function console.log.call with those parameters, which logically is equivalent to:

(function(a) {return a}).call(1,2)

This code produces the same result and is a little bit easier to understand - we are using call() on the unnamed function. call() 's parameters are a ' this ' object followed by arguments. In this case ' this ' is 1 , and the argument is 2 , so the function gets called with a assigned to 2 (the 1 is not used in the function which is unbound). So this will always return 2 , since it is simply returning the second item in the list.

But what is the role of alert , and why can we substitute any function name there? Well, it seems that there is a single call() function that is shared between all prototypes. You can verify that hypothesis by running

alert.call === console.log.call
true

So it doesn't matter if we use alert , console.log , or nothing, we are always using the same call() function.