AMDG Emil Dotchevski wrote:
Let's say I have a class template functor like this:
struct func { typedef void result_type;
template <class A1,class A2,class A3> void operator()( A1, A2, A3, int x ) { } };
Given a list of types, I want the function instantiated and called for all permutations of the types in the list; the additional argument(s), x, just need to be forwarded the same for all instances.
Actually I got it working, see http://codepad.org/tnYTy9Iw.
I don't think that my solution is very good though. First, it has some run-time overhead due to the use of boost::bind in fwd1 and fwd2 (see link above.) Second, I'd like the macro that defines the type list not to be hard-coded in fwd1 and fwd2.
Is this possible?
How does this work #include <boost/preprocessor/seq/for_each_product.hpp> #include <boost/preprocessor/seq/transform.hpp> #include <boost/preprocessor/seq/enum.hpp> #include <iostream> struct func { typedef void result_type; template <class A1,class A2,class A3> void operator()( A1, A2, A3, int x ) { std::cout << typeid(A1).name() << ',' << typeid(A2).name() << ',' << typeid(A3).name() << ',' << x << '\n'; } }; template<class T> T make() { return T(); } #define CALL_FUNC(r, args) func()(BOOST_PP_SEQ_ENUM(args), x); #define MAKE_OBJECT(s, data, elem) make<elem>() #define FUNC_II(types) \ BOOST_PP_SEQ_FOR_EACH_PRODUCT(CALL_FUNC, (types)(types)(types)) #define FUNC_I(types) FUNC_II(types) #define FUNC(types, arg) \ if(true) { \ int x = arg; \ FUNC_I(BOOST_PP_SEQ_TRANSFORM(MAKE_OBJECT, ~, types)) \ } else ((void)0) int main() { FUNC((int)(long), 42); } In Christ, Steven Watanabe