Firebug JS warning to jQuery 1.4.2 "The 'charCode' property of a keyup event should not be used. The value is meaningless." To ignore it?

Question

Firebug 1.5.4 JavaScript warning : The 'charCode' property of akeyupevent should not be used. The value is meaningless. To ignore it? Is there any issue?

The warning appears to jQuery 1.4.2 keyupandkeydown, not onkeypress.
I have read that on changingevent.keyCodeandevent.charCodetoevent.whichmust fix it, but it does not work for me.
Full code example in http://jsfiddle.net/zTevK/2/ and in question
My code useskeyupand does not work withkeypress.

  $(document).bind('keyup', function(e){
   var key = e.which;
   if (key > 36 && key < 41) {
    if (key == 37) { changeTab(-1); }
    if (key == 38) { changeTab(-imgPerRow); }
    if (key == 39) { changeTab(+1); }
    if (key == 40) { changeTab(+imgPerRow); }
    e.preventDefault();
  ...

Answer 1

This is a very common error message that comes out of mozilla code. The message comes out for every keystroke and it can lead to performance problems if you have the error console open. I've complained to mozilla about this and similar messages to no effect. jjb

Answer 2

The jQuery code itself normalizes every event in jQuery.event.fix

// props includes 'charCode' - this will access it
for ( var i = this.props.length, prop; i; ) {
  prop = this.props[ --i ];
  event[ prop ] = originalEvent[ prop ];
}

// also, later in the same function

// Add which for key events
if ( !event.which && ((event.charCode || event.charCode === 0) ? event.charCode : event.keyCode) ) {
  event.which = event.charCode || event.keyCode;
}

One of these lines of code is going to access charCode, which in turn creates your warning... You don't even need to do anything in your event handler (illustrated on jsfiddle)...

The "solution" I usually end up using is just running without JS warnings (errors still show up)

You can safely ignore this message (assuming you aren't using charCode, and are indeed using which)