Sorry, still don't understand.
As said, i have defined
void serialize(Archive & ar, MemberClass & m, const unsigned int version);

so
template<class Archive>
void serialize(Archive & ar, ContainerClass & c, const unsigned int version)
{
  ar & c.other_members; // easy - built-in type
  serialize(ar, c.special_member, version);  //use same version???
}
compiles.

My question is what *version* number do i need to provide in the call to serialize for c.special_member?
- The same as with what serialize(Archive & ar, ContainerClass & c, const unsigned int version) got called? (Like i did in example) This doesn't make sense to me.
- Or I need to handle it myself, say i need to call  serialize(ar, c.special_member, 3); because i *know* the current version of MemberClass is version 3.
- Or there is some archive magic i don't see.



On Mon, Mar 5, 2018 at 5:52 PM, Robert Ramey via Boost-users <boost-users@lists.boost.org> wrote:
On 3/5/18 8:36 AM, Lars Ruoff via Boost-users wrote:
Maybe i didn't express it clearly enough.
I want non-intrusive serialization.
I want to modify neither ContqinerClass, nor MemberClass.
(Both are just custom classes, no link with STL containers. I just wanted to express that Container contains an instance of Member)

So how can i write
ar & c.special_member;
since there is no serialize member function for this class?

 That's the problem.  the type of c.special_member must be serializable. If it's not already, you have to make it serializable by writing your own serialize function for it.  It can be either intrusive or non-intrusive - doesn't matter.  But it has to be there.  The versioning of type type of special_member is handled in the serialization of that type.


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