Hi Stephan, >>> ((void)(callFoo(bar) == good); >>>which you also trust the compiler to simplify. >>Honestly: no, I wouldn't. > Huh?
Ehm - if I would have known that OSL_VERIFY is expanded to this, then probably yes. Since I dind't know this, I wouldn't have truested it, since I never though about it. Something like this, book it as misunderstanding, sorry. >>expanded to >> callFoo(bar) == good >>There's probably a reason for this not working, which I overlook, is there? > > The cast to void is probably in OSL_VERIFY to not provoke compiler > warnings about unused code. Ah, thanks. Ciao Frank -- - Frank Schönheit, Software Engineer [EMAIL PROTECTED] - - Sun Microsystems http://www.sun.com/staroffice - - OpenOffice.org Database http://dba.openoffice.org - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
