how come #include "*.cpp" works?

[Dave Korn]

On 24 May 2006 17:00, mwoehlke wrote:

> 3: A create_<name> function must be written. If all objects were
> constructed the same, this could have been lumped into the declaration
> CREATABLE_CLASS macro, however not all constructors take the same
> arguments, which is why 'create_<name>' takes an argc/argv pair. IOW
> this requirement cannot be eliminated.

  Are you sure this can't be worked around using varargs macros?

#define CREATABLE_CLASS(name, ...) \
  extern class OBJECT* create_##name( __VA_ARGS__ ); \

> Thus, except for Makefile, it achieves the objective of having a list
> which is entirely generated and does not need to be maintained by hand.

> === For declaring an object
> #define CREATABLE_CLASS(name) \
> extern class OBJECT* create_##name( int argc, const char* argv[] ); \
> class name : virtual public OBJECT

> === So main.cpp doesn't need the headers for every class
> #undef CREATABLE_CLASS
> #define CREATABLE_CLASS(name) \
> extern class T_OBJ* create_##name( int argc, const char* argv[] );
> #include "objects.def"

  Bad idea, I think.  You now have two separate bits of code that are
generating the prototypes for this set of functions.  You run the danger of
getting inconsistent code.  In this case, unless T_OBJ is the exact same type
as OBJECT, you've already made (technically speaking) undefined behaviour,
haven't you?  Umm, perhaps I mean "Would someone remined me whether C++
mangled names include the return type, or just the args signature?"

  Perhaps it would be better to factor out the prototype generation into a
separate macro that could be invoked by both instances of "#define
CREATABLE_CLASS" ?

    cheers,
      DaveK
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