If you must serialize the variant then a serialize() will have to be defined for every type. Unless I have misunderstood what you mean by "really works": It's not about satisfying the compiler, but rather telling it how to serialize this member. By telling it (with an empty serialize() ) to not serialize or de-serialize anything, it works appropriately, as long as you read/write the same data in the same order (i.e. nothing :) ) Mit freundlichen Grüßen / With best regards Andreas Iwanowski - IT Administrator / Software Developer www.awato.de | namezero@afim.info T: +49 (0)2133 26031 55 | F: +49 (0)2133 26031 01 awato Software GmbH | Salm Reifferscheidt Allee 37 | D-41540 Dormagen avisor-Support | T: +49 (0)621 6094 043 | F: +49 (0)621 6071 447 Geschäftsführer: Ursula Iwanowski | HRB: Neuss 7208 | VAT-no.: DE 122796158 -----Original Message----- From: Boost-users [mailto:boost-users-bounces@lists.boost.org] On Behalf Of Merrill Cornish Sent: Sunday, 01 February, 2015 21:16 To: boost-users@lists.boost.org Subject: Re: [Boost-users] boost serialization + boost variant + boost blank If simply defining an empty serialize() function really works, then your suggestion is best. Next, question, does it really work--or just temporarily satisfy the compiler. _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users