generic way to access C++ libs?

Diez B. Roggisch deetsNOSPAM at web.de
Fri Nov 5 18:20:36 CET 2004


> What I would like to have is some module (or whatever), with which I can
> say "load this C++ library", and then, "create that C++ object" or "call
> method x of C++ object y".

C++ has no concept of runtime type information (except from saying "this is
of class X). This sort of information is known as "reflection" in java or
python, and that allows for late time bindings to those languages. In other
words: you can create on the fly wrappers for them. 

But  c++ lacking this means that you need to feed the specification of
objecst - the header files - to some generator. WHich is exactly what
happens in the wrapper generators like swig and sip. I never toyed around
with boost, but I don't believe they made the impossible possible.

So to answer your question: Its not possible. There are other reasons as
well: C++ defines no binary layout of objects, the result is that a lib
complied  with different compilers (gcc, intel, msvc) results in
incompatible binaries. So even if one would ship the lib with the header
files: to actually generate the wrappers, the appropriate compliler has to
be used. And as you rarely find compilers of different kinds on end user
machines, you'll have to do it on the developers machine.

> Without doing anything else (such as recompiling the library or generating
> wrappers).

As I just said: not doable.

Third thing is that differences in language design make it necessary to
invest developer time in creating wrappings: In c++ you can pass arguments
by a pointer to them, allowing to modify them in the callers stackframe (or
whereever they live). As this is not possible in python, you need to work
around this, usually by returning a tuple of modified values in addition to
the result of the method/function itself.

So there are strong limits on automatically generationg wrappers - a
minmimum of work has to be done.
-- 
Regards,

Diez B. Roggisch



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