实际上,解包操作可以一次定位两个不同的参数包(我认为它们需要的长度相等)。在这里,我们想要一组类型和一组数字。
一种近乎:
template <typename C, typename... T, size_t... N>
C* init_from_tuple_impl(bp::tuple tpl) {
return new C{ bp::extract<T>(tpl[N])... };
}
我们 “只是” 需要生成指数的包:
template <size_t... N> struct Collection {};
template <typename C> struct ExtendCollection;
template <size_t... N>
struct ExtendCollection< Collection<N...> > {
typedef Collection<N..., sizeof...(N)> type;
};
template <typename... T>
struct GenerateCollection;
template <>
struct GenerateCollection<> { typedef Collection<> type; };
template <typename T0, typename... T>
struct GenerateCollection<T0, T...> {
typedef typename ExtendCollection<
typename GenerateCollection<T...>::type
>::type type;
};
,然后使用它:
template <typename C, typename... T, size_t... N>
C* init_from_tuple_impl(bp::tuple tpl, Collection<N...>) {
return new C { bp::extract<T>(tpl[N])... };
}
template <typename C, typename... T>
C* init_from_tuple(bp::tuple tpl) {
typename GenerateCollection<T...>::type collection;
return init_from_tuple_impl<C, T...>(tpl, collection);
}
在行动,在Ideone 。
我们可以通过在init_from_tuple_impl
实施(除去new
为例)制作一个 “错误” 见证的正确性:
template <typename C, typename... T, size_t... N>
C* init_from_tuple_impl(bp::tuple tpl, Collection<N...>) {
return C { bp::extract<T>(tpl[N])... };
}
在行动,在Ideone:
prog.cpp: In function 'C* init_from_tuple_impl(bp::tuple, Collection<N ...>)
[with
C = bp::Object,
T = {int, float, char},
unsigned int ...N = {0u, 1u, 2u},
bp::tuple = std::basic_string<char>
]':
正是我们想要:)