How to implement a closure for this function in JavaScript?

Question

Last night, I Googled a lot and couldn't find the solution for my problem: I have a for loop with one function in it which gets me only the latest value from the array.

So, here is the example:

obj1.route = new Routeng();
obj2.route = new Routeng();

for(var x in arrObjs) { //arrObjs = array of objects
  var g = arrObjs[x];

  // I can access properties of all "g" objects

  Routelousse.gen(function(res) {
    var pathern = res.pathern;
    g.routel.staviPather(pathern);

    MYOBJ.vehicles.push(g);
    alert(g.name); // during the loop I always get the LAST "g" object from "arrObjs"
  }, g.point);

}

@annakata - Because JavaScript doesn't have block scope here, the same g is re-used every for loop pass, so the incorrect one is being used when that asynchronous function completes. — Nick Craver
– Nick Craver, Commented Nov 17, 2010 at 10:11
@Nick - yeah I understand block scope thanks, what I didn't understand is what the desired result was. But you did it seems so the customer's happy. — annakata
– annakata, Commented Nov 17, 2010 at 10:15
Are you sure you googled a lot? ;) JavaScript Closures for Dummies(Example 5) is quite helpful... — Felix Kling
– Felix Kling, Commented Nov 17, 2010 at 10:16

Nick Craver · Accepted Answer · 2010-11-17 10:07:51Z

2

It should look like this:

obj1.route = new Routeng();
obj2.route = new Routeng();

for(var x=0; x<arrObjs.length; x++) {
  var g = arrObjs[x];

  (function(ig) {
    Routelousse.gen(function(res) {
      var pathern = res.pathern;
      ig.routel.staviPather(pathern);

      MYOBJ.vehicles.push(ig);
      alert(ig.name);
    }, ig.point);
  })(g);
}

In this we're passing the current g into that self-executing function as a different variable, rather than the g which is shared in the function you're currently in (this isn't block scope) and is changing each pass of the for loop.

Also note the for loop change...you should never use a for...in loop to iterate an Array, use a normal for loop for that.

answered Nov 17, 2010 at 10:07

Nick Craver

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