[C++-sig] Static and non-static functions with the same name

[Nikolay Mladenov]

Please remind me, what happens if you do not put .staticmethod("garply"),
and then try :
 >>> Foo().garply(0,0)
 >>> Foo.garply("")

?

 >     class_<Foo>("Foo")
 >         .def("garply", (string (*)(string)) &Foo::garply)
 >         .def("garply", (string (Foo::*)(int, int)) &Foo::garply)
 >         .staticmethod("garply")


Bilokon, Paul wrote:
> Hi,
> 
> In C++ it's possible to have the following situation:
> 
> class Foo
> {
> public:
> 
>     string garply(int x, int y)
>     {
>         return "I am the non-static Foo::garply(int, int)";
>     }
> 
>     static string garply(string z)
>     {
>         return "I am the static Foo::garly(string)";
>     }
> };
> 
> Now I want to expose both the static and non-static garply to Python,
> and here I'm stuck:
> 
> using namespace boost::python;
> 
> BOOST_PYTHON_MODULE(overloading)
> {
>     class_<Foo>("Foo")
>         .def("garply", (string (*)(string)) &Foo::garply)
>         .def("garply", (string (Foo::*)(int, int)) &Foo::garply)
>         .staticmethod("garply")
>             // ??? Which garply is that ???
>     ;
> }
> 
> staticmethod takes a single string as its parameter, the function name.
> But I have both static and non-static variants. How do I specify which
> of them is static? [Technically it's possible. We already differentiate
> between overloaded functions on the Python side, so we just need to
> check if the first argument is "self", e.g. by looking at its type --
> hardly foolproof though, e.g. consider a "copy-constructor"
> signature...]
> 
> Of course, one may ask why don't I give the Python functions different
> names. But what if I'm trying to replicate the C++ API as faithfully as
> I can? I would like to avoid name changes.
> 
> What's the best thing to do?
> 
> Many thanks!
> 
> Regards,
> Paul
> 
> 
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