对于可以组成不同mixin类型的类型,请考虑以下代码。复合类型的构造函数采用可变的元组列表,这些元组表示组合类型的构造函数的参数:
#include <string>
#include <tuple>
#include <utility>
struct MixinBase {
MixinBase() = default;
// Note; want to delete these instead of default them.
MixinBase(const MixinBase&) = default;
MixinBase(MixinBase&&) = default;
};
struct MixinA : public MixinBase {
MixinA(int, const std::string&, const std::string&) {}
};
struct MixinB : public MixinBase {
MixinB(const std::string&, const std::string&) {}
};
template <typename... Mixins>
struct Composition : private Mixins... {
template <typename... Packs>
Composition(Packs&&... packs)
: Mixins(constructMixin<Mixins>(
std::forward<Packs>(packs),
std::make_index_sequence<std::tuple_size_v<Packs>>{}))...
{
}
private:
template <typename Mixin, typename Pack, size_t... Indexes>
Mixin constructMixin(Pack&& arguments, std::index_sequence<Indexes...>) const
{
return Mixin(std::get<Indexes>(std::forward<Pack>(arguments))...);
}
};
int main()
{
std::string a{"a"};
std::string b{"b"};
Composition<MixinA, MixinB>(
std::forward_as_tuple(7, a, b), std::forward_as_tuple(a, b));
return 0;
}
这工作得很好,但是,我想避免通过constructMixin
进行间接访问,而是直接构造每个继承的mixin对象,以便可以避免在mixin类型上使用复制/移动构造方法。这可能吗?
答案 0 :(得分:4)
您可以定义一个帮助器类来支持分段构造:
template <typename T>
struct Piecewise_construct_wrapper : T {
template <typename Tuple>
Piecewise_construct_wrapper(Tuple&& t) :
Piecewise_construct_wrapper(std::forward<Tuple>(t),
std::make_index_sequence<std::tuple_size_v<Tuple>>{}) {}
template <typename Tuple, size_t... Indexes>
Piecewise_construct_wrapper(Tuple&& t, std::index_sequence<Indexes...>) :
T(std::get<Indexes>(std::forward<Tuple>(t))...) {}
};
然后,您可以让Composition
继承Piecewise_construct_wrapper<Mixins>...
:
template <typename... Mixins>
struct Composition : private Piecewise_construct_wrapper<Mixins>... {
template <typename... Packs>
Composition(Packs&&... packs)
: Piecewise_construct_wrapper<Mixins>(std::forward<Packs>(packs))...
{
}
};
答案 1 :(得分:3)
我想避免通过constructMixin进行间接操作,而是直接构造每个继承的mixin对象,以便可以避免在mixin类型上使用复制/移动构造方法。这可能吗?
您将需要Mixins选择加入以直接允许分段构建。不幸的是,这是非常重复的,因此您可以使用如下宏:
#define PIECEWISE_CONSTRUCT(Type) \
template <typename Tuple> \
Type(std::piecewise_construct_t, Tuple&& tuple) \
: Type(std::piecewise_construct, \
std::forward<Tuple>(tuple), \
std::make_index_sequence<std::tuple_size_v<Tuple>>()) \
{ } \
template <typename Tuple, size_t... Indexes> \
Type(std::piecewise_construct_t, Tuple&& tuple, \
std::index_sequence<Indexes...>) \
: Type(std::get<Indexes>(std::forward<Tuple>(tuple))...) \
{ }
用作:
struct MixinA : public MixinBase {
MixinA(int, const std::string&, const std::string&) {}
PIECEWISE_CONSTRUCT(MixinA)
};
struct MixinB : public MixinBase {
MixinB(const std::string&, const std::string&) {}
PIECEWISE_CONSTRUCT(MixinB)
};
template <typename... Mixins>
struct Composition : private Mixins... {
template <typename... Packs>
Composition(Packs&&... packs)
: Mixins(std::piecewise_construct, std::forward<Packs>(packs))...
{ }
};
不幸的是,保证复制省略不能在构造子对象时起作用-因此,这可能是最好的选择。我认为如果没有某种嵌套包,您是否可以直接进行此操作?我很有可能只是没有足够的创造力,如果有人提出更好的建议,我会很好奇。