Re: [boost] [Serliazation] Auto registration/physical coupling (was: Thread-safety again)

From: Emil Dotchevski (emildotchevski_at_[hidden])
Date: 2007-01-04 17:05:40

> Sorry, I don't think that's an accurate statement of anything.  This
> is accurate:
>
>   In a given translation unit (TU), all combinations of A and T,
>   where A is an archive registered (in the TU) with
>   BOOST_SERIALIZATION_REGISTER_ARCHIVE and T is a class exported (in
>   the TU) with BOOST_CLASS_EXPORT(**) are used to instantiate
>   polymorphic pointer serialization code.  Period.
>
>   The translation unit or some portion of it may later be discarded
>   if the linker can prove none of its functions or objects are used,
>   or else its initializers may be skipped.  Separate issue.

Just to clarify, besides the portability issues, I brought up the issue of 
the templates being instantiated by the BOOST_CLASS_EXPORT macro because it 
introduces physical coupling between the classes being registered and the 
serialization library. Perhaps I need to clarify what I mean.

//foo.h

class foo;

template <class A>
void operator<<( A &, foo const & );

class foo
{
public:
  foo();
private:
  int m_;
  template <class A>
  friend void operator<<( A &, foo const & );
};

template <class A>
void operator<<( A & a, foo const & x )
{
  a << x.m_;
}

The above foo.h defines class foo, and also specifies how objects of class 
foo are serialized, yet there is no (physical) coupling between foo.h and a 
serialization library, because foo.h does not include any headers. This is a 
good thing: typically, most of the compilation units that #include "foo.h" 
will not serialize foo objects; the ones that do serialize foo objects will 
know to #include "foo.h" and a (compatible) serialization library. Also 
consider that I can write a program that makes use of class foo yet doesn't 
use/include/link a serialization library.

This would not be possible if foo.cpp auto-registers class foo with a 
serialization library.