'Base class' shared libraries in C?

[Dennis Melentyev]

Hi,

Yes, if you'll not pretend it to be even a "small C++".

IMHO, the simplest way to get "overloaded methods" is to implement base 
class methods pointers table, some kind of derived "classes" 
initialization (registering overrided methods, setting base class 
pointers in children, etc). Actualy, it's the way it implemented in C++.

Anyway, you framework should take a BIG care of probability to call 
"pure virtual" analogs (uninitialized or not overloaded members).

In your particular case, using of multiple modules would be impossible 
or inconvenient, because of each module would contain the same code for 
"base class", confusing linker/rtld.

Isn't it easier to explicitly link to "base" module and at least one 
"derived" module?
Or, to use just normal C++?
The later will also give you a compiler driven "spellchecker" for free
:)

Jonathon McKitrick wrote:
Hi all,

I am trying to implement some C++ functionality in C.

Suppose I have several libraries that share behavior and properties, such as
version, working directory, and so on.  I want other libraries to extend
this behavior without having to duplicate any code.  For instance:
typedef struct base
{
	char version[32];
	char path[32];
} base_t;
typedef struct foo
{
	base_t *base;
	int data;
} foo_t;
typedef struct bar
{
	base_t *base;
	int data;
} bar_t;
lib-base.so:
	base_get_version(base_t *, char *);
	base_get_path(base_t *, char *);
lib-derived-foo.so:
	foo_one_method(foo_t *, int data);
	foo_two_method(foo_t *, int data);
lib-derived-bar.so:
	bar_one_method(bar_t *, int data);
	bar_two_method(bar_t *, int data);
I want the derived modules to link with the base library, so that my
application can link with only the derived modules and still be able to call
the base functions without having to link to that library or call 'wrapper'
functions in the derived modules.
Is there a way to do this?

jm


--
Dennis Melentyev
http://www.melentyev.kiev.ua/dennis