The existing answers test for very specific properties of std::map
, either that it is precisely a specialization of std::map
(which would be false for std::unordered_map
or non-standard types with the same interface as std::map
), or testing that its value_type
is exactly std::pair<const key_type, mapped_type>
(which would be true for multimap
and unordered_map
, but false for non-standard types with similar interfaces).
This only tests that it provides key_type
and mapped_type
members, and can be accessed with operator[]
, so doesn't say that std::multimap
is mappish:
#include <type_traits>
namespace detail {
// Needed for some older versions of GCC
template<typename...>
struct voider { using type = void; };
// std::void_t will be part of C++17, but until then define it ourselves:
template<typename... T>
using void_t = typename voider<T...>::type;
template<typename T, typename U = void>
struct is_mappish_impl : std::false_type { };
template<typename T>
struct is_mappish_impl<T, void_t<typename T::key_type,
typename T::mapped_type,
decltype(std::declval<T&>()[std::declval<const typename T::key_type&>()])>>
: std::true_type { };
}
template<typename T>
struct is_mappish : detail::is_mappish_impl<T>::type { };
Because is_mappish
has a "base characteristic" of either true_type
or false_type
you can dispatch on it like so:
template <typename T>
auto foo(const T& items, true_type)
{
// here be maps
}
template <typename T>
auto foo(const T& items, false_type)
{
// map-free zone
}
template <typename T>
auto foo(const T& items)
{
return foo(items, is_mappish<T>{});
}
Or you can avoid dispatching entirely, and just overload foo
for maps and non-maps:
template <typename T,
std::enable_if_t<is_mappish<T>{}, int> = 0>
auto foo(const T& items)
{
// here be maps
}
template <typename T,
std::enable_if_t<!is_mappish<T>{}, int> = 0>
auto foo(const T& items)
{
// map-free zone
}
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