Assigning Random Values to Object Elements

Question

I am assigning random values between 1-50 to object elements, I have 5 objects, I don't know why but all the objects are getting the same random values...

here is my code :

var SmileyRed = {
   radius: 15,
   xspeed: 0,
   yspeed: 0,
   xpos:350,  // x-position of smiley
   ypos: 65  // y-position of smiley
};

var SmileyReds = new Array();

 for (var i=0; i<5; i++){
 SmileyReds[i] = SmileyRed;
 SmileyReds[i].xspeed = Math.floor((Math.random()*50)+1);
 SmileyReds[i].yspeed = Math.floor((Math.random()*50)+1);
 }

SmileyReds[0].xspeed and SmileyReds[3].xspeed have the same values but shouldn't they be different?

Your array contains 5 references to the same object. There is only one SmileyRed. — bfavaretto
– bfavaretto, Commented Nov 10, 2013 at 20:10
@bfavaretto I just figured it out the original SmileyRed shows also the same random values, I thought I was making a copy of SmileyRed each time... how can I make them different independent objects? — Anarkie
– Anarkie, Commented Nov 10, 2013 at 20:13

charlietfl · Accepted Answer · 2013-11-10 20:19:58Z

3

The problem is that when you make an object equal to another object, the new object is a refrence of the original, not a copy.

What is happening is you are creating 5 referencs to the original SmileyRed. Essentially when you change one, you change all. So only value being applied in your loop is from last pass of the loop, the previous 4 passes are overwritten.

You could change to:

var SmileyReds = new Array();

 for (var i=0; i<5; i++){
/* new object each pass*/
 SmileyReds[i] =  {
   radius: 15,
   xspeed: 0,
   yspeed: 0,
   xpos:350,  // x-position of smiley
   ypos: 65  // y-position of smiley
};
 SmileyReds[i].xspeed = Math.floor((Math.random()*50)+1);
 SmileyReds[i].yspeed = Math.floor((Math.random()*50)+1);
 }

Another way would be:

var SmileyRed = function(){
    return{
       radius: 15,
       xspeed: 0,
       yspeed: 0,
       xpos:350,  // x-position of smiley
       ypos: 65  // y-position of smiley
    };
}

 for (var i=0; i<5; i++){
    /* new object each pass*/
     SmileyReds[i] =  SmileyRed();/* note () this time*/

edited Nov 10, 2013 at 20:19

answered Nov 10, 2013 at 20:13

charlietfl

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4 Comments

Anarkie Over a year ago

I see this is creating 5 objects, isn't there a way to copy SmileyRed and then work on the elements? I think copying this object would be easier.

Pointy Over a year ago

@Anarkie there's no built-in object copy mechanism in JavaScript. You can write a function to do it, but it's a somewhat complicated problem in general.

XCS Over a year ago

Maybe by using jQuery.extend api.jquery.com/jQuery.extend stackoverflow.com/questions/122102/…

charlietfl Over a year ago

yes...see new update...I know you need to create more at various intervals in your game

Peter Rasmussen · Accepted Answer · 2013-11-10 20:22:39Z

3

The problem is that the indexes from 0 to 4 contains references to the same object SmileyRed. You should create a new object for each iteration if you want to seperate them.

So you are actually changing the same object in each iteration. Thereby you will always use the last random numbers (from the last object).

By calling a function which returns an object, you'll get a new object for every iteration. Like below.

var SmileyRed = function() {
    return {
       radius: 15,
       xspeed: 0,
       yspeed: 0,
       xpos:350,  // x-position of smiley
       ypos: 65  // y-position of smiley
    }
};

var SmileyReds = new Array();

 for (var i=0; i<5; i++){
     SmileyReds[i] = SmileyRed();
     SmileyReds[i].xspeed = Math.floor((Math.random()*50)+1);
     SmileyReds[i].yspeed = Math.floor((Math.random()*50)+1);
 }

JSfiddle

edited Nov 10, 2013 at 20:22

answered Nov 10, 2013 at 20:12

Peter Rasmussen

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1 Comment

Pointy Over a year ago

In this case, because of the way you wrote the function, you don't even need new.

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