如何在可变参数模板中有多个参数包？

Question

函数one() 接受一个参数包。函数 two() 接受两个。每个包都被限制在 A 和 B 类型中。为什么无法实例化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.

Answer 1

这是另一种使用模板模板参数拥有多个参数包的方法：

#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;
}

Answer 2

我找到了一种解决方案。将每个参数包包装在一个元组中。使用结构进行部分特化。这是一个演示，它通过将一个元组作为列表使用并累积另一个来将参数转发给函子。好吧，这个通过复制转发。类型推导使用元组，函数参数中没有使用元组，我认为这很整洁。

#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间列表。

Answer 3

函数模板（如skypjack的例子）和类和变量模板的部分特化可以有多个参数包，如果模板参数包之后的每个模板参数要么有一个默认值，要么可以推导出来。我想添加/指出的唯一一件事是，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;
};      

Answer 4

编译器需要一种方法来知道两个可变参数模板之间的障碍在哪里。一种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参数模板。 （将最后一级保留为函数）

#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;
}