Victor,
Thanks for putting that clearer (and more correctly) than I did!
Object#extend does return the extended object, but also affects the
first parameter ("destination") in-place, extending it without
requiring the return.
My initial example actually re-defined the base class in terms o
superclass's constructor in prototype. So may anyone know how to use the "bind" function? I was wondering if it is anything related to inheritance. Merry Christmas and happy coding to all of you. Regards, QnA 2005/12/25, Cliff Mees <[EMAIL PROTECTED]>: Here's one approac
Here's one approach:Labour = Class.create();Labour.prototype = { initialize: function(name) { this.name = name; }, showMeYourName: function() { alert(this.name); }}Worker = Class.create();Worker.prototype = Object.extend(Labour.prototype, { initialize: function(name, position) { this.