call, apply, bind

Call

function a(){
    console.log('a');
}

// These two are the same
a() === a.call()

Questions

Example

var b = {
    name: 'jay'
    say(): {
        console.log(this)
    }
}

var c = {
    name: 'jay'
    say(): {
        return function() {
            console.log(this)
        }
    }
}

var d = {
    name: 'jay'
    say(): {
        return () => console.log(this)
    }
}

// What would each return?
b.say() 

c.say()()

d.say()()

Answer

var b = {
    name: 'jay'
    say(): {
        console.log(this)
    }
}
b.say()

b.say() would return {name: "jay", say: fn()} Because the say() function is "bounded" to the b object

var c = {
    name: 'jay'
    say(): {
        return function() {
            console.log(this)
        }
    }
}
c.say()()

This would return window that's because the function isn't bounded to anything. As in it's . is not called by anything.

c.say() is run, and it'd equal something like var temp = c.say() and then temp()would then be executed. And since temp isn't bounded to anything it'd just be the window object.

var d = {
    name: 'jay'
    say(): {
        return () => console.log(this)
    }
}
d.say()()

This would return {name: 'jay', say: fn} Although it might look the same as c the main difference is that arrow function is lexically scopes the function