A question about call in javascript

One Question about Call

Question:

function fn1() {
    console.log(1);
};
function fn2() {
    console.log(2);
};
fn3 = fn2.call;
fn2.call(fn1);
fn3.call(fn1);

Please determine what we get?

Solution:

2
1

Key Points: * fn1.call is a function reference. which is: * Function.prototype.call = fn3 = fn1.call; * When fn1.call() is running, what is this inside the call? Notice the this inside the call and this inside the fn1 are totally different. The this inside call is fn1. The one who called call method, it is the this of call method. On the other hand, this inside call is the one who called call method. * Call is also a method, it can use the call method also. * The this in the class method, it is like this is calling the method.

Thus,

fn3.call(fn1) == fn2.call(fn1) = Function.prototype.call.call(fn1) = fn1.call() = fn1; Thus, we get 1 as a result.