qua*_*ant 3 c++ templates constructor variadic-templates c++11
假设我有一个基类,以后可以通过派生它来"扩展",让我们调用这个类Base和扩展Derived.类的模板签名是固定的,不能更改(即我们不能更改类的模板参数).Derived该类的作者一无所知Base,只是它的构造函数可能需要一些参数.
但是,最终派生类的调用者知道应该传递多少个参数.我怎么写这个Derived扩展名?这是我有的:
struct Base
{
Base(int baseArg) {}
};
struct Derived : public Base
{
template <typename... Args>
Derived(Args&&... args, int derivedArg)
:
Base(std::forward<Args>(args)...)
{
}
};
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当我尝试运行它时,Derived d(1, 1);我得到以下erorr消息:
prog.cpp: In function 'int main()':
prog.cpp:19:16: error: no matching function for call to 'Derived::Derived(int, int)'
Derived d(1, 1);
^
prog.cpp:19:16: note: candidates are:
prog.cpp:11:2: note: template<class ... Args> Derived::Derived(Args&& ..., int)
Derived(Args&&... args, int myArg)
^
prog.cpp:11:2: note: template argument deduction/substitution failed:
prog.cpp:19:16: note: candidate expects 1 argument, 2 provided
Derived d(1, 1);
^
prog.cpp:8:8: note: constexpr Derived::Derived(const Derived&)
struct Derived : public Base
^
prog.cpp:8:8: note: candidate expects 1 argument, 2 provided
prog.cpp:8:8: note: constexpr Derived::Derived(Derived&&)
prog.cpp:8:8: note: candidate expects 1 argument, 2 provided
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构造函数Derived应该使用2个参数,使用第一个构造自身并将第二个传递给基类.为什么这不起作用?
nth 获取一些参数的第n个元素:
template<size_t n, class...Args>
auto nth( Args&&... args )
->typename std::tuple_element<n,std::tuple<Args&&...>>::type
{
return std::get<n>( std::forward_as_tuple(std::forward<Args>(args)...) );
}
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这使用上面的方法来提取最后一个,但是所有但最后的参数,并将它们发送到适当的位置:
struct Derived : public Base {
struct tag{};
template <typename... Args>
Derived(Args&&... args) : Derived(
tag{},
std::make_index_sequence<sizeof...(Args)-1>{},
std::forward<Args>(args)...
){}
template<size_t...Is, class...Args>
Derived(tag, std::index_sequence<Is...>, Args&&...args ):
Base(nth<Is>(std::forward<Args>(args)...)...)
{
int derivedArg = nth<sizeof...(Args)-1>(std::forward<Args>(args)...);
}
};
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我们为前n-1个元素构建一个序列,将它们传递给base,并为自己存储最后一个元素.
但是,如果你先把额外的论点放在首位,那就容易多了.