Harvey Chapman wrote: > On Oct 30, 2009, at 1:35 PM, Matthias Melcher wrote: >> so as long as there is one window visible, run() will not return. >> >> How about: >> >> while (myWindow->shown()) Fl::run(); > > That is almost the same as Fl::run() by itself. I want: > > while (myWindow (or any window) exists, shown or hidden) Fl::run();
There's no way how FLTK can determine, if any window "exists, shown or hidden". That's something you must do yourself. Honestly, I still don't understand what you want to achieve. If any window (all windows) would exist, but also be hidden, then there's no way for user interaction, and thus Fl::run() is useless (and that's the reason why it exits). Everything else is up: to you to determine, if your program should continue running and eventually show() any of its windows (again). Then you can call Fl::run() again (see my other post). Albrecht _______________________________________________ fltk mailing list [email protected] http://lists.easysw.com/mailman/listinfo/fltk
