Reputation: 2442
I have this code:
auto myTuple = hana::tuple_t<int, char*, long>;
std::cout << boost::typeindex::type_id<decltype(myTuple[1_c])>().pretty_name() << std::endl;
This outputs:
boost::hana::type_impl<char*>::_
I want to access the 'char*' type, but if I do:
std::cout << boost::typeindex::type_id<decltype(myTuple[1_c])::type>().pretty_name() << std::endl;
It outputs:
error: 'decltype(myTuple[1_c])' (aka 'boost::hana::type_impl<char *>::_ &') is not a class, namespace, or scoped enumeration
std::cout << boost::typeindex::type_id<decltype(myTuple[1_c])::type>().pretty_name() << std::endl
It's because it is a reference, if I do:
std::cout << boost::typeindex::type_id<decltype(boost::hana::traits::remove_reference(myTuple[1_c]))::type>().pretty_name() << std::endl;
Then it outputs 'char*'.
Is this the way to access the types of a tuple_t? There must be less cumbersome way.
Upvotes: 1
Views: 156
Reputation: 3093
This is indeed tricky. Hana provides a unary plus operator on hana::type
that decays any ref-qualified hana::type
to a rvalue. So basically,
#include <boost/hana.hpp>
namespace hana = boost::hana;
using namespace hana::literals;
auto myTuple = hana::tuple_t<int, char*, long>;
using T = decltype(+myTuple[1_c])::type;
// ^~~~~ notice unary plus here
Also note that you might be interested in using hana::experimental::print
from <boost/hana/experimental/printable.hpp>
. This is an experimental (hence unstable) feature, but I can assure you it should eventually make its way into the library, in one form or the other:
#include <boost/hana.hpp>
#include <boost/hana/experimental/printable.hpp>
#include <iostream>
namespace hana = boost::hana;
int main() {
auto myTuple = hana::tuple_t<int, char*, long>;
std::cout << hana::experimental::print(myTuple) << std::endl;
}
Output:
(type<int>, type<char*>, type<long>)
Edit: The unary plus operator is documented in the reference of hana::type
.
Upvotes: 4