> > You can do: > > my.cool.A.prototype.foo.call(this);
Alternatively, here is a variant that avoids the superclass' class name: this.constructor.superclass.prototype.foo.apply(this, arguments); (See Playground sample http://tinyurl.com/38ae7y9) But this is generally frowned upon (Why did you override foo in my.cool.B if you then try to get around it?!). T. ------------------------------------------------------------------------------ ThinkGeek and WIRED's GeekDad team up for the Ultimate GeekDad Father's Day Giveaway. ONE MASSIVE PRIZE to the lucky parental unit. See the prize list and enter to win: http://p.sf.net/sfu/thinkgeek-promo _______________________________________________ qooxdoo-devel mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/qooxdoo-devel
