The parent function is addEvent() and the child function is the one created by this line:
wrapper = function(e) {
handler.call(context, e);
}

When wrapper function is called, it will have access to all of the local variables in addEvent (element, eventName, handler, context, itself) but it will only use handler and context. This is because Javascript creates a closure:
http://helephant.com/2008/10/javascript-closures/

var n = function() and function n() result in exactly the same thing, a new function called n created in the current context. The difference is the first one is an operation while the second is a statement. You can use the function operator wherever it is valid to use a Javascript expression (in a loop, as a parameter to a function, as a return expression) so it’s a lot more flexible. If you don’t need the flexibility of the function operator, you can use either syntax.

There’s more information in the anonymous function post:
http://helephant.com/2008/08/javascript-anonymous-functions/

I’ve rewritten the two sections about event handlers to hopefully make them a bit clearer and given them better example. Hope this helps.

Javascript has a feature called closures so when you create an anonymous function inside another function, you can access local variables from the parent function. We can use this to reference the function context object when the anonymous function is run.

You have to click on full example to figure out what handler is. I almost thought it was the button (as I’m new at this) and not some function with the same name.

What’s “e” for?

You say you need a closure… What are the local variables and what is the parent function. Is the parent function? As there are many functions in your example, which one are you referring to with “function context object”.

Finally I noticed you’ve used var n = function() and function n(). What is the difference between the two constructs?

Have a look at http://connecteu.hieruwwebsite.nl/ if you want, that’s what I’ve been working on.

Thanks!

The browser works from last to first when painting the div so the last one is at the back and the first one is at the front. That’s why the last one is the only one it makes sense to have a background colour on. If the browser painted a background colour when processing one of the other background declarations it would overwrite all the backgrounds that were underneath it.

I’m not sure what you mean by IE8. What’s not working?

I’ve also applied a background-color to the divs with multiple backgrounds. It seems that you can’t use this in a one-line statement, such as:
background: #fff url(/image1.png), url(/image2.png);
Is this correct? To make it work, I have to add the background-color after I set the background-images.

All works fine in FF and IE7, but in IE8 only the background-color is shown. Might there be a solution to this?

Best wishes,

Wil

I think it’s particularly helpful if you are using a Javascript library because many of them use prototypes to simulate classic inheritance and if things go wrong it’s really hard to debug without understanding what’s going on.

That said.. I should update my article with a more convincing argument in the object prototype’s favour. :)

The Pet.prototype.sayHello example is simplistic to illustrate the idea. In the real world, javascript classes are much more significant in size, and can have rather large methods. If I have 1000 instances of an object, without prototype I would have 1000 instances of that large method as well. With prototype, the sayHello method will be included but only as a reference to the single existing method, which translates to a significant reduction in memory usage.
I’m all for simplified programming, but would have to defend prototype as a necessity.

public class PieControl1 : Control
{
…
public int Pies
{
get { return (int) (ViewState["Pies"] ?? 0); }
set { ViewState["Pies"] = value; }
}
}

The new ?? operator says, “Use the value if it is not null, or else use the second operand if it is.” So if ViewState["Pies"] is null, a 0 will be returned. Otherwise, the value of ViewState["Pies"] will be returned.