why is my javascript object method returning undefined?

Question

I'm a beginner in both javascript oop and game programming too(!) . Here i've created a game player with a method. but the method is returning undefined. why is that?,

bobsGame = {};

bobsGame.player  = function(which){
  this.which = which;
  this.rollDice = function () {
    diceVal = Math.floor(Math.random() * 6 + 1);
    console.log(diceVal);
    return diceVal;
  }
}

var player1 = new bobsGame.player('player1');

and then in the markup…

$('#roll-dice-btn-1').click(function(){
  bobsGame.player1.rollDice();
});

There are too many incoherent things in here. Where is defined bobsgame.player1 for example ? Can you build a working fiddle demonstrating your problem ? — Denys Séguret
– Denys Séguret, Commented Jan 9, 2014 at 16:03

adeneo · Accepted Answer · 2014-01-09 16:05:09Z

3

There is no bobsGame.player1 in your class, you just instanciated a new instance to the variable player1 ?

var player1 = new bobsGame.player('player1');

$('#roll-dice-btn-1').click(function(){
    player1.rollDice();
});

FIDDLE

answered Jan 9, 2014 at 16:05

adeneo

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3 Comments

egl Over a year ago

This. You aren't currently calling the rollDice() method on the player1 object you created.

adeneo Over a year ago

@egl - Then what am I calling ?

egl Over a year ago

Your answer is fine, was referring to the OP calling the method incorrectly.

palerdot · Accepted Answer · 2014-01-09 16:05:08Z

2

Your call should be player1.rollDice() like

  $('#roll-dice-btn-1').click(function(){
    player1.rollDice();
  });

You are confused between player property of bobsGame object and the player1 object you actually created.

answered Jan 9, 2014 at 16:05

palerdot

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Comments

azurelogic · Accepted Answer · 2014-01-09 16:05:33Z

2

this might work a little better

bobsGame.player1 = new bobsGame.player('player1');

answered Jan 9, 2014 at 16:05

azurelogic

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2 Comments

adeneo Over a year ago

And why would you instanciate the class to a property of itself ?

Alex Shilman Over a year ago

Or just replace bobsGame completely

Lawren · Accepted Answer · 2014-01-09 16:22:26Z

1

bobsGame.player1 is undefined because player1 was not declared within the object bobsGame.

If you want to create player1 as a key in your bobsGame object, you must use bobsGame.player1 = new bobsGame.player('player1'); instead. So your code would look like: bobsGame = {};

bobsGame.player  = function(which){
  this.which = which;
  this.rollDice = function () {
    diceVal = Math.floor(Math.random() * 6 + 1);
    console.log(diceVal);
    return diceVal;
  }
}

var player1 = new bobsGame.player('player1');

Otherwise, you can use:

$('#roll-dice-btn-1').click(function(){
  player1.rollDice();
});

answered Jan 9, 2014 at 16:22

Lawren

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Comments

Louis XIV · Accepted Answer · 2014-01-09 16:08:57Z

1

You can play with a structure like that:

bobsGame = function(opts) {
    this.player = opts.player;
    this.diceVal = 0;
}

bobsGame.prototype.rollDice  = function(){
    this.diceVal = Math.floor(Math.random() * 6 + 1);
}

And in your main file:

var player1 = new bobsGame({player: 'player1'});
$('#roll-dice-btn-1').click(function(){
    player1.rollDice();
    console.log(player1.diceVal);
});

JsFiddle

answered Jan 9, 2014 at 16:08

Louis XIV

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Comments

Grish · Accepted Answer · 2014-01-09 16:16:51Z

1

slight problem with your. It cannot find bobsGame.player1 in your class. So change like this and it would work fine..

bobsGame.player  = function(which){
    this.which = which;

    this.rollDice = function(){
        diceVal = Math.floor(Math.random() * 6 + 1);
        console.log(diceVal);
        return diceVal;
    }
}


     var player1 = new bobsGame.player('player1');
    player1.rollDice();

answered Jan 9, 2014 at 16:16

Grish

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