This seems to be a very simple question: How does one remove the first (the n-th) type in a std::tuple
?
Example:
typedef std::tuple<int, short, double> tuple1;
typedef std::tuple<short, double> tuple2;
The operation described above would transform tuple1
into tuple2
. Is it possible?
You can use a simple type function based on partial specialization of a class template:
#include <type_traits>
#include <tuple>
using namespace std;
template<typename T>
struct remove_first_type
{
};
template<typename T, typename... Ts>
struct remove_first_type<tuple<T, Ts...>>
{
typedef tuple<Ts...> type;
};
int main()
{
typedef tuple<int, bool, double> my_tuple;
typedef remove_first_type<my_tuple>::type my_tuple_wo_first_type;
static_assert(
is_same<my_tuple_wo_first_type, tuple<bool, double>>::value,
"Error!"
);
}
Also, this solution can be easily generalized to remove the i-th type of a tuple:
#include <type_traits>
#include <tuple>
using namespace std;
template<size_t I, typename T>
struct remove_ith_type
{
};
template<typename T, typename... Ts>
struct remove_ith_type<0, tuple<T, Ts...>>
{
typedef tuple<Ts...> type;
};
template<size_t I, typename T, typename... Ts>
struct remove_ith_type<I, tuple<T, Ts...>>
{
typedef decltype(
tuple_cat(
declval<tuple<T>>(),
declval<typename remove_ith_type<I - 1, tuple<Ts...>>::type>()
)
) type;
};
int main()
{
typedef tuple<int, bool, double> my_tuple;
typedef remove_ith_type<1, my_tuple>::type my_tuple_wo_2nd_type;
static_assert(
is_same<my_tuple_wo_2nd_type, tuple<int, double>>::value,
"Error!"
);
}