Why is a function wrap unside a function object not called in javascript [duplicate]

Question

I have JavaScript function as object:

function hexMesh(){
     var side=25;
     console.log('hexMesh');

     function drawOver(){
     }
}

As you can see it has a function call drawOver.

I try to call it by using a constructor as follows:

window.onload = function() {
    hexMeshObj=new hexMesh();
    hexMeshObj.drawOver();
}

But it gives me error saying undefined is not a function

Now I know I could declare the function in the prototype of the object but I don't want to do that.

Here it is on JSFiddle.

Community · Accepted Answer · 2017-05-23 12:03:50Z

1

You can't use JavaScript like that!

Solution: Use Class-like prototypes (please don't treat them as classes, although they provide inheritance)

var x = function(v){ this.test = v; } // This is your constructor
x.prototype.show = function(){ console.log("TEST", this.test); } // This is a prototype with this as a context

var y = new x(true);
y.show(); // Logs TEST true

Edit: Alternatively (although the prototype way is better as it provides real inheritance the oop way)

var x = function(v){
var context = this;
this.test = v;
this.show = function(){
    console.log('test', context.test)
});

Another alternative way is to use bind to bind context if you need it.

edited May 23, 2017 at 12:03

CommunityBot

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answered Nov 7, 2015 at 23:11

Schahriar SaffarShargh

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4 Comments

forgivenson Over a year ago

It might be helpful to include some info as to why using prototypes is better than the second way.

Schahriar SaffarShargh Over a year ago

Added reference. Thanks for the suggestion.

Snedden27 Over a year ago

I was looking for the alternate way, I knew I could do it using prototype but I didn't want the function to be in the prototype of the object as it doesn't make sense in my case, Thanks

Schahriar SaffarShargh Over a year ago

I added info regarding bind. You can achieve the same context (that's the extent of it) using bind(this). On a second note this question is not a duplicate as it does not ask for oop strictly in the manner that the referenced question is set.

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