Calling a javascript function as an array method

Question

Update: I know there are existing implementations of forEach method. The purpose of doing this is to learn and improve my Javascript skills.

I am trying to implement a forEach method for array objects. What I am doing is the following:

var list = ['one', 'two', 'three'];

function forEach(callback){
    for(var n = 0; n < this.length; n++){
        callback.call(this[n], n);
    }
};

list.forEach(function(index){
        console.log(index);
        console.log(this);
    }
);

I am not very good with javascript and I am trying to get better so I've been reading a little bit and I now know that if I do this kind of thing, the context of the "forEach" function would be the object which is calling it, in this case "list".

When this code runs I get the error: "Uncaught TypeError: Object one,two,three has no method 'forEach'".

What is it I am not understanding?

Thanks!

You did nothing to associate the function forEach with the forEach property of your list object. — Raymond Chen
– Raymond Chen, Commented Aug 6, 2012 at 20:55

Joe · Accepted Answer · 2012-08-06 20:54:19Z

1

var list = ['one', 'two', 'three'];

list.forEach = function(callback){
    for(var n = 0; n < this.length; n++){
        callback.call(this[n], n);
    }
};

list.forEach(function(index){
        console.log(index);
        console.log(this);
    }
);

It seems you are trying to make your own on that list object. If so you need to make it a property of that object.

answered Aug 6, 2012 at 20:54

Joe

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Comments

Billy Moon · Accepted Answer · 2012-08-06 21:01:18Z

I think what you want to do is probably not to attach the foreach function to only the one array you have here, but to make it work for all arrays. To do that, you must edit the Array prototype (something that some people have very strong opinions about, because you can not protect against potential future namespace collisions - but other people find extremely useful). Once the Array.prototype object has a function added to it, you can call that function on any array. Like this:

var list = ['one', 'two', 'three'];

Array.prototype.myForEach = function(callback){
    for(var n = 0; n < this.length; n++){
        callback.call(this[n], n);
    }
};

list.myForEach(function(index){
    console.log(index);
    console.log(this);
});

*N.B. to avoid conflict with existing forEach functions (https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/forEach) I have named the function myForEach which I expect to be safe from conflict.

Aaron Kurtzhals · Accepted Answer · 2012-08-06 20:59:44Z

0

I believe that you will need to define forEach before you instantiate the array.

Array.prototype.forEach = function (callback) {
    var n;
    for(n = 0; n < this.length; n++){
        callback.call(this[n], n);
    }
};

answered Aug 6, 2012 at 20:59

Aaron Kurtzhals

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1 Comment

Bergi Over a year ago

No, you don't need to. JavaScript is dynamic.

gaspyr · Accepted Answer · 2012-08-06 21:02:34Z

0

You need to add the forEach method to the prototype of Arrays :

/*
* pay attention to the fact that if you use forEach aas method name you'll override
* the built default forEach method, try to use some prefix to distinguish between
* the built in methods and your own ones 
*/
Array.prototype.my_forEach = function(callback){ /* your code here */ };

answered Aug 6, 2012 at 21:02

gaspyr

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Comments

Bergi · Accepted Answer · 2012-08-06 21:02:54Z

0

You could set the context object (this keyword) by using call:

myForEach.call(list, function(index) { … });

An other possibility would be to call the function as a property of the list object:

list.myForEach = function(callback) { … };
list.myForEach(function(index) { … });

or enable this by adding it to the prototype of all Array objects:

Array.prototype.myForEach = function(callback) { … };
['one', 'two', 'three'].myForEach(function(index) { … });

Note: there should already be a forEach method on Arrays. Use the compatibilty shim from there for ancient browsers to provide exactly the same functionality.

edited Aug 6, 2012 at 21:02

answered Aug 6, 2012 at 20:57

Bergi

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