[Boost-users] Re: Guide to using serialization macro BOOST_CLASS_EXPORT

From: Bill Lear (rael_at_[hidden])
Date: 2005-01-07 14:54:05

On Friday, January 7, 2005 at 10:58:13 (-0800) Robert Ramey writes:
>...
>The whole concept behind BOOST_CLASS_EXPORT has been problematic for me for
>a couple of reasons.  As a result, the confusion you cite is a real one.

I settled on the following that a friend suggested, which seems to
work.

In my serialization/util.h file, I have the following:

  template <class T>
  void save(const T* t, const std::string& filename) { ... }

  template <class T>
  T* load(const std::string& filename) { ... }

at the bottom of this file, I have the following:

  namespace dummy { struct bogus { static int bogus_method(); }; }

  // force linking with util.cc
  static int bogus_variable = dummy::bogus::bogus_method();

in serialization/util.cc, I have:

  #include <serialization/util.h>

  // expensive include which we wish to relegate to one one source file.
  #include <AllDerived.h>

  // the trigger method that forces others to link with this module
  int dummy::bogus::bogus_method() { return 0;}

The AllDerived.h file has includes for all my headers and all the
BOOST_CLASS_EXPORT macros.

In my application code, I do this:

  #include <Base.h>
  #include <serialization/util.h>

and it all works flawlessly.  I am spared from including the headers
for my derived classes, and the accompanying export macros.  The save
and load routines work as expected.

Bill