what does this usage of apply() means in Javascript

Please, can someone tell me what does this.init.apply(this, arguments) do in the code below?

I understand what apply() does in general, but in the context of the code below, what is it doing in there?

var Class = function() {

    var klass = function() {
        this.init.apply(this, arguments); //I don't really get this bit...
    };

    klass.prototype.init = function(){};

    return klass;
};

var Person = new Class;

//Usage
var someone =  new Person;

I see a lot of people using it. I've got an idea of what it does but can't really put my hands on it so I need more light.

I'm going up an extra level in JS, so I wanna know everything about it, not just the simple 'Hello world' level.

Many thanks

Answers:

Answer

apply is a member function of a function object. Suppose we have:

function saySomething(thing, anotherThing) {
    alert(this + " says " + thing + " and " + anotherThing);
}

Then we can use:

saySomething.apply(document, ["hello", "goodbye"]);

This calls the function and supplies the values of the array as arguments to the function. The first argument species the context of the function (or, what this equals when the function runs).

You should also be aware that arguments is a special variable that holds an array of all arguments passed to the function. So, here, this.init.apply(this, arguments) means that the init function is called and passed all the of arguments that were passed to the klass constructor.

In a real implementation, I think init would expect arguments. Consider how this would be done without apply:

var klass = function(arg1, arg2, arg3) {
    init(arg1, arg2, arg3);
}

klass.prototype.init = function(arg1, arg2, arg3) {}

If you wanted to add arg4 to init, you'd have add it in three places! By having klass transparently pass all its arguments to init, you make you code much less brittle.

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.