如何在一个可变模板中拥有多个参数包

How can I have multiple parameter packs in a variadic template?

本文关键字:拥有 参数 一个      更新时间:2023-10-16

函数one()接受一个参数包。函数two()接受两个。每个包都被限制为封装在类型AB中。为什么不可能实例化two()

template <typename T>
struct A {};
template <typename T>
struct B {};
template <typename... Ts>
void one(A<Ts> ...as) {
}
template <typename... Ts, typename... Us>
void two(A<Ts> ...as, B<Us> ...bs) {
}
int main() {
  auto a = A<int>();
  auto b = B<int>();
  // Just fine
  one();
  one(a);
  one(a, a);
  // All errors    
  two();
  two(a);
  two(a, b);
}

尝试使用gcc和clang。

sam@wish:~/x/cpp$ gcc -std=c++0x variadic_templates.cpp 
variadic_templates.cpp: In function ‘int main()’:
variadic_templates.cpp:23:7: error: no matching function for call to ‘two()’
variadic_templates.cpp:23:7: note: candidate is:
variadic_templates.cpp:11:6: note: template<class ... Ts, class ... Us> void two(A<Ts>..., B<Us>...)
variadic_templates.cpp:24:8: error: no matching function for call to ‘two(A<int>&)’
variadic_templates.cpp:24:8: note: candidate is:
variadic_templates.cpp:11:6: note: template<class ... Ts, class ... Us> void two(A<Ts>..., B<Us>...)
variadic_templates.cpp:25:11: error: no matching function for call to ‘two(A<int>&, B<int>&)’
variadic_templates.cpp:25:11: note: candidate is:
variadic_templates.cpp:11:6: note: template<class ... Ts, class ... Us> void two(A<Ts>..., B<Us>...)
sam@wish:~/x/cpp$ clang -std=c++0x variadic_templates.cpp 
variadic_templates.cpp:23:3: error: no matching function for call to 'two'
  two();
  ^~~
variadic_templates.cpp:11:6: note: candidate function template not viable: requires at least 1 argument, but 0 were provided                                                                                                                 
void two(A<Ts> ...as, B<Us> ...bs) {}
     ^
variadic_templates.cpp:24:3: error: no matching function for call to 'two'                                                                                                                                                                   
  two(a);
  ^~~
variadic_templates.cpp:11:6: note: candidate function not viable: requires 0 arguments, but 1 was provided                                                                                                                                   
void two(A<Ts> ...as, B<Us> ...bs) {}
     ^
variadic_templates.cpp:25:3: error: no matching function for call to 'two'                                                                                                                                                                   
  two(a, b);
  ^~~
variadic_templates.cpp:11:6: note: candidate function not viable: requires 0 arguments, but 2 were provided                                                                                                                                  
void two(A<Ts> ...as, B<Us> ...bs) {}
     ^
3 errors generated.

以下是使用模板模板参数获得多个参数包的另一种方法:

#include <iostream>
template <typename... Types>
struct foo {};
template < typename... Types1, template <typename...> class T
         , typename... Types2, template <typename...> class V
         , typename U >
void
bar(const T<Types1...>&, const V<Types2...>&, const U& u)
{
  std::cout << sizeof...(Types1) << std::endl;
  std::cout << sizeof...(Types2) << std::endl;
  std::cout << u << std::endl;
}
int
main()
{
  foo<char, int, float> f1;
  foo<char, int> f2;
  bar(f1, f2, 9);
  return 0;
}

我找到了一个解决方案。将每个参数包包裹在一个元组中。使用结构进行部分专门化。这里有一个演示,通过将一个元组作为列表并累积另一个元组,将参数转发到函子。好吧,这个是通过复制转发的。元组用于类型推导,但函数参数中没有使用元组,我认为这很巧妙。

#include <iostream>
#include <tuple>
template < typename ... >
struct two_impl {};
// Base case
template < typename F,
           typename ...Bs >
struct two_impl < F, std::tuple <>, std::tuple< Bs... > >  {
  void operator()(F f, Bs... bs) {
    f(bs...);
  }
};
// Recursive case
template < typename F,
           typename A,
           typename ...As,
           typename ...Bs >
struct two_impl < F, std::tuple< A, As... >, std::tuple< Bs...> >  {
  void operator()(F f, A a, As... as, Bs... bs) {
    auto impl = two_impl < F, std::tuple < As... >, std::tuple < Bs..., A> >();
    impl(f, as..., bs..., a);
  }
};
template < typename F, typename ...Ts >
void two(F f, Ts ...ts) {
  auto impl = two_impl< F, std::tuple < Ts... >, std::tuple <> >();
  impl(f, ts...);
}
struct Test {
  void operator()(int i, float f, double d) {
    std::cout << i << std::endl << f << std::endl << d << std::endl;
  }
};
int main () {
  two(Test(), 1, 1.5f, 2.1);
}

元组是一个非常好的编译时列表。

如果模板参数包后面的每个模板参数都有默认值或可以推导,则函数模板(如skypjack的示例)以及类和变量模板的部分专门化可以具有多个参数包。我唯一想补充/指出的是,对于类和变量模板,您需要部分专业化。(参见:C++模板,完整指南,Vandevoorde,Josuttis,Gregor 12.2.4,第二版)

// A template to hold a parameter pack
template < typename... >
struct Typelist {};
// Declaration of a template
template< typename TypeListOne 
        , typename TypeListTwo
        > 
struct SomeStruct;
// Specialization of template with multiple parameter packs
template< typename... TypesOne 
        , typename... TypesTwo
        >
struct SomeStruct< Typelist < TypesOne... >
                 , Typelist < TypesTwo... >
                 >
{
        // Can use TypesOne... and TypesTwo... how ever
        // you want here. For example:
        typedef std::tuple< TypesOne... > TupleTypeOne;
        typedef std::tuple< TypesTwo... > TupleTypeTwo;
};      

编译器需要一种方法来知道两个可变模板之间的屏障在哪里。一个干净的方法是为对象定义一组参数,为静态成员函数定义第二组参数。这可以通过在彼此中嵌套多个结构来应用于两个以上的可变模板。(保持最后一级功能)

#include <iostream>
template<typename... First>
struct Obj
{
    template<typename... Second>
    static void Func()
    {
        std::cout << sizeof...(First) << std::endl;
        std::cout << sizeof...(Second) << std::endl;
    }
};
int main()
{
    Obj<char, char>::Func<char, char, char, char>();
    return 0;
}
two<>();
two<int>(a);
two<int>(a, b);

参见https://www.youtube.com/watch?v=va9I2qivBOA29:00-35:00