这是可能的,但并非微不足道。为了使这个工作,一个包含N
参数的元组必须支持2^N
构造函数,所有的组合T&&
和T const&
for each T
。
我们要做的是混合这些2^N
构造函数,这些构造函数可以使用继承来完成。由于基类的构造函数只能using
显式使用,因此我们只能添加固定数量的基类的构造函数,因此我们必须使用递归。
一种方法是从0
to计数,2^N
如果第 i 位为 1,则使第 i 个参数为 const-ref,否则为右值。总共使用基类执行此操作,2^N
其中每个基类都向其直接基类添加一个构造函数。
namespace detail {
// A bitlist holds N powers of two: 1, 2, 4, 8, 16, ...
template <std::size_t... i> struct bitlist { using type = bitlist; };
template <std::size_t N, typename=bitlist<>>
struct make_bitlist;
template <std::size_t N, std::size_t... i>
struct make_bitlist<N, bitlist<i...>>
: make_bitlist<N-1, bitlist<0,1+i...>> {};
template <std::size_t... i> struct make_bitlist<0, bitlist<i...>>
: bitlist<(1<<i)...> {};
struct forward_tag {}; // internal struct that nobody else should use
// if T is a reference, some constructors may be defined twice, so use a non-accessible type.
template <bool B, typename T>
using const_if_set = typename std::conditional<B,
typename std::conditional<std::is_reference<T>::value, forward_tag, T const&>::type, T&&>::type;
// Our helper class. Each tuple_constructor is derived from N-1 others
// each providing one constructor. N shall equal (1<<sizeof...(T))-1
template <std::size_t N, typename L, typename... T> struct tuple_constructor;
template <std::size_t N, std::size_t... I, typename... T>
struct tuple_constructor<N, bitlist<I...>, T...>
: tuple_constructor<N-1, bitlist<I...>, T...>
{ // inherit base constructors
using tuple_constructor<N-1, bitlist<I...>, T...>::tuple_constructor;
tuple_constructor(const_if_set<(N & I), T>... t)
: tuple_constructor<N-1, bitlist<I...>, T...>
(forward_tag{}, std::forward<const_if_set<(N & I), T>>(t)...) {}
};
// base case: N=0, we finally derive from std::tuple<T...>
template <std::size_t... I, typename... T>
struct tuple_constructor<0, bitlist<I...>, T...> : std::tuple<T...> {
tuple_constructor(T&&... t)
: tuple_constructor(forward_tag{}, std::forward<T&&>(t)...) {}
// All constructor calls are forwarded to this one
template <typename... T2>
tuple_constructor(forward_tag, T2&&... t2)
: std::tuple<T...>(std::forward<T2>(t2)...) {}
};
// Convenience using for N=2^n, bitlist=1,2,4,...,2^n where n = sizeof...(T)
template <typename... T>
using better_tuple_base = tuple_constructor
< (1<<sizeof...(T)) - 1, typename make_bitlist<sizeof...(T)>::type, T... >;
}
template <typename... T> struct better_tuple : detail::better_tuple_base<T...> {
using typename detail::better_tuple_base<T...>::tuple_constructor;
};
住在 LWS
但请注意,这不能很好地扩展到大型元组并显着增加编译时间。在我看来,这是语言的限制。