使用 boost.python 导出 c++ 多态函数的智能方法

Question

I have a highly polymorphic c++ function that can be called with nearly 20 diferent types.

In order to expose it to python I'm doing something like this:

#include originalFunctionNamespace.hpp

template<class T>
    T foo(T x)
    {
        return orignalFunctionNamespace::foo(x);
    }

but then to call them from python I have to specialize the functions with each type they support:

BOOST_PYTHON_MODULE(module_foo)
{
    def("foo", foo<orignalFunctionNamespace::type1>);
    def("foo", foo<orignalFunctionNamespace::type2>);
    def("foo", foo<orignalFunctionNamespace::type3>);
    def("foo", foo<orignalFunctionNamespace::type4>);
    def("foo", foo<orignalFunctionNamespace::type5>);
    def("foo", foo<orignalFunctionNamespace::type6>);
    def("foo", foo<orignalFunctionNamespace::type7>);
    ...
    ...
    ...
    def("foo", foo<orignalFunctionNamespace::typeN>);
}

This works, but I can't stop thinking that there must be a smarter way to do it. Since I have to do it for many many functions, things are getting big and super repetitive. Any suggestions?

Answer 1

Looks like a possible use for typelist magic, to me.

If you're using (or can use) the new C++0x standard, you can write something like this:

template <typename ArgType>
void def_foo_overloads()
{
    def("foo", foo<ArgType>);
    // last type in list
}
template <typename ArgType, typename... MoreArgTypes>
void def_foo_overloads()
{
    def("foo", foo<ArgType>);
    def_foo_overloads(MoreArgTypes...);
}

// use it like:
def_foo_overloads<orignalFunctionNamespace::type1,
                  orignalFunctionNamespace::type2,
                  ...
                  orignalFunctionNamespace::typeN> ();

If you don't have access to the new standard, you can do the equivalent thing using the old recursive typelists.