Question

函数 one（）接受一个参数包。函数 two（）接受两个。每个包都被约束为包含在 A 和 B 类型中。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.

Answer 1

以下是使用模板模板参数包含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;
}

Answer 2

我找到了一个解决方案。将每个参数包包装在元组中。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元组来将参数转发给仿函数。好吧，这个通过复制转发。元组在类型推导中使用但在函数参数中没有使用元组，我认为它是整齐的。

#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

如果模板参数包之后的每个模板参数具有默认值或可以推导，则

功能模板（如skypjack的示例）以及类和变量模板的部分专业化可以具有多个参数包。我唯一要添加/指出的是，对于类和变量模板，您需要部分专业化。（请参阅：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模板之间的障碍。一种干净的方法是为一个对象定义一组参数，为静态成员函数定义第二组参数。通过相互嵌套多个结构，可以将其应用于两个以上的可变参数模板。（保持最后一级的功能）

#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;
}

如何在可变参数模板中包含多个参数包？

4 个答案: