CODEWITHSUNDEEP
javascriptfunctionparameterscall
Rodrigo Puente
Rodrigo Puente

Reputation: 101

Javascript call function

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?

Upvotes: 3

Views: 5343

Answers (4)

vibhu
vibhu

Reputation: 377

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']);

Upvotes: 0

Guy Sirton
Guy Sirton

Reputation: 8401

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.

Upvotes: 5

Zero Fiber
Zero Fiber

Reputation: 4465

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().

Upvotes: 2

jfriend00
jfriend00

Reputation: 707158

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().

Upvotes: 1

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