Question

这个问题的灵感来自我之前的问题No template parameter deduction of parameter pack。

请考虑以下代码示例：

#include <memory>
#include <string>

template<typename... FArgs>
class Callback
{
    public:
    class Handle{};
};

class BaseCallbackHandle
{
};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template<typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle( H handle)
{
   return {};
}

int main()
{
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h); //should compile fine
    makeTypeErasedCallbackHandle(s); //should raise a compile error
}

另见http://coliru.stacked-crooked.com/a/5f2a2e816eef6afd

函数模板makeTypeErasedCallbackHandle现在将任何类作为输入参数。有没有办法确保（例如使用static-assert或enable_if），只允许Callback<FArgs...>::Handle（任何FArgs）作为H？ Callback<int>::Handle的示例应该编译，而std::string将失败。

Answer 1

一种方法是通过传递一些额外的参数：

template <typename... Pack> struct packer {};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template <typename... T>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle(typename Callback<T...>::Handle h, T...)
{
   return {};
}

template <typename... T>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle_2(typename Callback<T...>::Handle h, packer<T...>)
{
   return {};
}

int main()
{
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h, 0); //should compile fine
    // OR
    makeTypeErasedCallbackHandle_2(h, packer<int>());

    //makeTypeErasedCallbackHandle(s); //should raise a compile error
}

这使用identity trick（Stephan T. Lavavej）来进行类型演绎。

Answer 2

在Handle类中定义类型，并在makeTypeErasedCallbackHandle()中引用该类型：

#include <memory>
#include <string>

template <typename... FArgs>
struct Callback {
    struct Handle {
        using callback_type = Callback<FArgs...>;    
    };
};

struct BaseCallbackHandle {
};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template <typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle(H handle) {
    using callback_type = typename H::callback_type;

    return {};
}

int main() {
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h); //should compile fine
    makeTypeErasedCallbackHandle(s); //should raise a compile error
}

Live example

在实例化期间，对于没有定义嵌套类型的任何H，这将失败。

通过更多的努力，您可以static_assert向客户端生成有意义的消息，同时通过类型特征增加解决方案的灵活性。这样做的好处是callback_impl::is_callback可以专门用于任意句柄类型：

#include <memory>
#include <string>

namespace callback_impl {

struct callback_identification_type {};

template <typename T, typename = void>
struct is_callback : std::false_type {};

template <typename T>
struct is_callback<T,
    std::enable_if_t<std::is_same<typename T::callback_id_type,
                     callback_identification_type>::value>>
 : std::true_type {};

}

template <typename... FArgs>
struct Callback {
    struct Handle {
        using callback_id_type = callback_impl::callback_identification_type;    
    };
};

struct BaseCallbackHandle {
};

using TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>;

template <typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle(H handle) {
    static_assert(callback_impl::is_callback<H>::value,
                  "The handle type is not a member of a recognised Callback<T...>");
    return {};
}

int main() {
    Callback<int>::Handle h;
    std::string s;
    makeTypeErasedCallbackHandle(h); //should compile fine
    makeTypeErasedCallbackHandle(s); //should raise a compile error

    return 0;
}

Live example

输出：

g++ -std=c++14 -O2 -Wall -Wno-unused-local-typedefs -pedantic -pthread main.cpp && ./a.out
main.cpp: In instantiation of 'TypeErasedCallbackHandle makeTypeErasedCallbackHandle(H) [with H = std::__cxx11::basic_string<char>; TypeErasedCallbackHandle = std::unique_ptr<BaseCallbackHandle>]':
main.cpp:41:35:   required from here
main.cpp:32:5: error: static assertion failed: The handle type is not a member of a recognised Callback<T...>
     static_assert(callback_impl::is_callback<H>::value,
     ^~~~~~~~~~~~~

Answer 3

您可以为您的班级成员制作虚构成员，并在替代方面将其称为： http://coliru.stacked-crooked.com/a/d5738766fd7ac45f

class Handle
{
    public:
    static void doNothing(){}
};

...

TypeErasedCallbackHandle makeTypeErasedCallbackHandle( H handle)
{
   H::doNothing();
   return {};

更标准的方法是使用类特征或静态断言，来自像BOOST这样的库（可能其中一些已经是语言的一部分）。

编辑： http://coliru.stacked-crooked.com/a/2a3adcb9d9dd274c 这种方式稍好一些，var而不是call。

Answer 4

class BaseCallback
{
  ~BaseCallback() = 0;
};

template<typename... FArgs>
class Callback : public BaseCallback
{
  ~Callback(){...}
  ...
};

template<typename H>
TypeErasedCallbackHandle makeTypeErasedCallbackHandle( H handle)
{
  static_assert(std::is_base_of<BaseCallback, H>::value, "Must use a callback");
  return {};
}

此应有效，只允许您在回调中使用该功能

仅为特定的模板化类启用模板

4 个答案: