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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.


class foo;

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

class foo
  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.

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