I have some template code which takes a shared pointer to a class and call a function or method. The problem comes up, if the called method is defined as const
.
Example:
struct Y {};
struct X
{
const Y Go() const { return Y{}; }
const Y Go2() { return Y{}; }
};
Y f1( std::shared_ptr<X> ) { return Y{}; }
template< typename FUNC, typename ... ARGS >
auto Do( std::shared_ptr<X>& ptr, FUNC&& f, ARGS&& ... args )
{
return f( ptr, std::forward<ARGS>(args)... );
}
template < typename CLASS, typename RET, typename ... ARGS>
auto Do( std::shared_ptr<X>& base_ptr, RET (CLASS::*mem_ptr)( ARGS...), ARGS&& ... args )->RET
{
return (base_ptr.get()->*mem_ptr)( std::forward<ARGS>(args)...);
}
// Any chance to avoid the full duplication of the code here
// to define the member pointer to a const method?
template < typename CLASS, typename RET, typename ... ARGS>
auto Do( std::shared_ptr<X>& base_ptr, RET (CLASS::*mem_ptr)( ARGS...) const, ARGS&& ... args )->RET
{
return (base_ptr.get()->*mem_ptr)( std::forward<ARGS>(args)...);
}
int main()
{
auto xptr = std::make_shared<X>();
Y y1 = Do( xptr, &X::Go );
Y y2 = Do( xptr, &X::Go2 );
Y y3 = Do( xptr, &f1 );
}
My problem is the last specialization with the RET (CLASS::*mem_ptr)( ARGS...) const
. I simply want to stop duplication the whole code only for the const. In real world code the function calls again another templated one, which results in duplicating a lot of code here.
Any chance to get rid of the specialization for the const member pointer?
You might do:
template< typename FUNC, typename ... ARGS >
auto Do( std::shared_ptr<X>& ptr, FUNC&& f, ARGS&& ... args )
-> decltype((f(ptr, std::forward<ARGS>(args)... )))
{
return f( ptr, std::forward<ARGS>(args)... );
}
template<typename MemberF, typename ... ARGS>
auto Do(std::shared_ptr<X>& base_ptr, MemberF mem_ptr, ARGS&& ... args)
-> decltype((base_ptr.get()->*mem_ptr)( std::forward<ARGS>(args)...))
{
return (base_ptr.get()->*mem_ptr)( std::forward<ARGS>(args)...);
}