Javascript call function

Question

I have been testing some code lately trying to understand javascript a little bit better. Then I came across the call() function wich I can't get to understand well.

I have the following code:

function hi(){
    console.log("hi");
}

var bye = function(param, param2){
    console.log(param);
    console.log(param2);
    console.log("bye");
}

If I call bye.call(hi(), 1, 2), I get hi 1 2 undefined

And if I call bye.cal(1,2), I get 2 undefined bye undefined

for which I understand the call() function first parameter has to be a function, followed by the amount of parameter my bye function accepts. But where is the last undefined coming from?

Guy Sirton · Accepted Answer · 2013-06-10 05:15:20Z

5

This first parameter doesn't have to be a function. The first parameter is the object to which the "this" variable is set to in the context of the function call.

var bye = function(param, param2){
    console.log(param);
    console.log(param2);
    console.log("bye");
    console.log(this.x)
}

t = {'x': 1};

bye.call(t, 1, 2);

And the console should show: 1, 2, "bye" and 1.

The undefined is the return value of your function.

In your first call:

bye.call(hi(), 1, 2)

You're calling hi() (so it prints 'hi'), the return value is not used, and 1 and 2 are the parameters to bye.

In your second call:

bye.cal(1,2)

1 is assigned to this. 2 is param, and param2 is undefined.

edited Jun 10, 2013 at 5:15

answered Jun 10, 2013 at 5:08

Guy Sirton

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Comments

jfriend00 · Accepted Answer · 2013-06-10 05:12:12Z

2

You are getting the undefined because you function does not return anything, it only prints output to the screen. So, your code could be like this:

var obj = {foo: "hi"};
var bye = function(param, param2){
    console.log(this.foo);
    console.log(param);
    console.log(param2);
}

bye.call(obj, 1, 2)   // "hi", 1, 2

You can read here at MDN for more info on .call().

edited Jun 10, 2013 at 5:12

jfriend00

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answered Jun 10, 2013 at 5:06

Zero Fiber

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Comments

jfriend00 · Accepted Answer · 2013-06-10 05:09:44Z

1

fn.call() allows you to set the value that this will have when the function is called. That value of this must be the first argument to fn.call().

answered Jun 10, 2013 at 5:09

jfriend00

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Comments

vibhu · Accepted Answer · 2020-12-03 12:43:59Z

call method required all the parameter individually apply method required all parameter in an array explained with example.

let solarSystem = {
    sun: 'red',
    moon : 'white',
    sunmoon : function(){
       let dayNight = this.sun + ' is the sun color and present in day and '+this.moon + ' is the moon color and prenet in night';
        return dayNight;
    }
}

let work = function(work,sleep){
    console.log(this.sunmoon()); 
    // accessing the solatSystem it show error undefine sunmmon untill now because we can't access directly for that we use .bind(),.call(),.apply()
    console.log('i work in '+ work +' and sleep in '+sleep);
}

let outPut = work.call(solarSystem,'day','night');
let outPut1 = work.call(solarSystem,['day1','night1']);
let outPut2 = work.apply(solarSystem,['day2','night2']);

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