是否有一个c ++特性可以在C++中找到两种类型之间最受限制的类型？

Question

我想要一个类型特征 common

以便

common<int,int>::type              -> int
common<const int, int>::type       -> const int
common<int, int &>::type           -> int
common<int &, int &>::type         -> int &
common<int &, int const &>::type   -> int const &

那就是结果类型应该是两者中更受限制的.在C++ 11标准中是否有一个可以做到这一点的特性,还是我必须自己动手？

我的用例是我有类似的东西

template <typename T0, typename T1>
struct Foo {

   BOOST_STATIC_ASSERT(
    std::is_same
    < typename std::decay<T0>::type
    , typename std::decay<T1>::type
    >::value
   );

   // I need to find T which is the most restrictive common
   // type between T0 and T1
   typedef typename common<T0,T1>::type T

   T0 t0;
   T1 t1;

   T choose(bool c){
       return c ? t0 : t1;
   }

} 

Answer 1

恐怕你需要自己动手。您可以在 std::tuple 中扭曲您的类型，然后将其传递给std::common_type，例如

#include <tuple>
#include <type_traits>

template <class T1, class T2>
struct common {
    using type = typename std::tuple_element<0, typename std::common_type<std::tuple<T1>, std::tuple<T2>>::type>::type;
};

template <class T>
struct common<const T, T> {
    using type = const T;
};

template <class T>
struct common<T, const T> {
    using type = const T;
};

template <class T>
struct common<const T, const T> {
    using type = const T;
};

int main()
{
    static_assert(std::is_same<common<int, int>::type, int>::value, "");
    static_assert(std::is_same<common<const int, int>::type, const int>::value, "");
    static_assert(std::is_same<common<int, int &>::type, int>::value, "");
    static_assert(std::is_same<common<int &, int &>::type, int &>::value, "");
    static_assert(std::is_same<common<int &, int const &>::type, int const &>::value, "");
    return 0;
}

但你必须为const.