在 C 中包装 C++ 成员函数 - Visual Studio 2013 模板问题

Question

我正在创建一个轻量级的跨平台插件框架，它在应用程序和插件之间使用 C 接口（通常，但并非总是用 C++ 编写）。

要帮助 C++ 应用程序和插件编写者，我面临的一个挑战是找到一种简单的方法来跨 C 接口公开 C++ 对象功能。我目前的解决方案感觉很简单，并使用模板来“构建”包装底层 C++ 成员函数的 C 签名函数，基于 this great stackoverflow question and answer

template <typename Tc, typename F, F>
struct MemberFuncWrapper;

template <typename Tc,              // C interface structure tag
          typename T,               // C++ class, derived from Tc
          typename R,               // C++ member function return type
          typename ...Args,         // C++ member function argument types
          R (T::*f)(Args...) const> // C++ member function
struct MemberFuncWrapper<Tc, R (T::*)(Args...) const, f> {
  static R call(const Tc * tc, Args... args) {
    const T * t = static_cast<const T *>(tc);
    return ((*t).*f)(args...);
  }
};

此模板的实例化在 linux (gcc) 和 mac (clang) 下编译和运行良好，但在 Visual Studio 2013 中编译失败：

error C2440: 'specialization' : cannot convert from 'overloaded-function' to 'void (__cdecl Greeter::* )(void) const'
error C2973: 'MemberFuncWrapper<Tc,R(__cdecl T::* )(Args...) const,f>' : invalid template argument 'overloaded-function'

下面的独立示例代码显示了 Visual Studio 失败的行（在Greeter 类定义中）。我希望有人可以：

• 提供一种可以安抚 Visual Studio 的解决方法，或者
• 提出实现类似目标的替代策略

这里的独立代码演示了在相当冗长的Hello world 应用程序中使用 C++ 类实现 C 接口的上下文中使用的模板代码：

#include <iostream>
#include <utility>

//
// C interface and function(s) typically defined elsewhere
//
#ifdef __cplusplus
extern "C" {
#endif

  // The C interface implemented by a 'greeter'
  struct greeter_c {
    void(*greet_cb)(const struct greeter_c * greeter,
                    const char * recipient);
  };

  // Some C function that makes use of a greeter
  void broadcast(const struct greeter_c * greeter) {
    greeter->greet_cb(greeter, "world");
  }

#ifdef __cplusplus
}  // extern "C"
#endif


//
// Template magic that envelopes a C++ member
// function call in a C-signature function
//
template <typename Tc, typename F, F>
struct MemberFuncWrapper;

template <typename Tc,              // C interface structure tag
          typename T,               // C++ class, derived from Tc
          typename R,               // C++ member function return type
          typename ...Args,         // C++ member function argument types
          R (T::*f)(Args...) const> // C++ member function
struct MemberFuncWrapper<Tc, R (T::*)(Args...) const, f> {
  static R call(const Tc * tc, Args... args) {
    // Cast C structure to C++ object
    const T * t = static_cast<const T *>(tc);

    // Details such as catching/handling exceptions omitted.

    // Call C++ member function
    return ((*t).*f)(args...);
  }
};

// Repeat of the above for non-const member functions omitted


//
// A C++ class that implements the C 'greeter' interface
//
class Greeter : public greeter_c {
 public:
  // Constructor
  Greeter(const char * greeting) : m_greeting(greeting) {
    // Set up C interface callback by wrapping member function

    // !! The following line causes the Visual Studio compilation error !!
    greet_cb = MemberFuncWrapper<greeter_c,
                                 void (Greeter::*)(const char *) const,
                                 &Greeter::greet>::call;
  }

  // C++ member function that 'does' the greeting
  void greet(const char * recipient) const {
    std::cout << m_greeting << " " << recipient << std::endl;
  }

 private:
  const char * m_greeting;
};


// An application that greets using a Greeter's C interface
int main(int argc, char * argv[]) {
  // Create C++ object that implements C interface
  Greeter a("Hello");

  // Greet using Greeter's C interface
  broadcast(&a);

  return 0;
}

技术细节：

• Linux 开发：Centos 7、g++ (GCC) 4.8.3 20140911
• Mac 开发：Apple LLVM 版本 6.1.0 (clang-602.0.49)
• Windows 开发：Visual Studio Express 2013 Update 5 CTP

Answer 1

前言： std::forward 在这里没用，因为Args... 是明确指定的。换句话说，实例化的 C 接口不再是模板。 std::forward 在非模板代码中没有用处。因此，std::forward 未用于以下解决方案。

版本 1：

template <typename Base, typename Derived, typename R, typename... Args>
struct c_interface_gen {
  template <R(Derived::*mem_fn)(Args...)> inline
  static R invoke(Base* pb, Args... args) {
    return (static_cast<Derived*>(pb)->*mem_fn)(args...);
  }
  template <R(Derived::*mem_fn)(Args...) const> inline
  static R invoke(const Base* pb, Args... args) {
      return (static_cast<const Derived*>(pb)->*mem_fn)(args...);
  }
};

此版本有效。但这绝不是优雅的。主要问题在于使用该工具的语法冗长且不直观。

版本 2：

template <typename Sig>
struct mem_fn_sig;

template <typename R, typename D, typename... Args>
struct mem_fn_sig<R(D::*)(Args...)> {
  template <R(D::*mem_fn)(Args...)>
  struct mem_fn_inst {
    template <typename Base>
    struct base {
      inline static R invoke(Base* pb, Args... args) {
        return (static_cast<D*>(pb)->*mem_fn)(args...);
      }
    };
  };
};

template <typename R, typename D, typename... Args>
struct mem_fn_sig<R(D::*)(Args...) const> {
  template <R(D::*mem_fn)(Args...) const>
  struct mem_fn_inst {
    template <typename Base>
    struct base {
      inline static R invoke(const Base* pb, Args... args) {
        return (static_cast<const D*>(pb)->*mem_fn)(args...);
      }
    };
  };
};

template <typename Sig, Sig inst, typename Base>
struct c_interface_gen:
  mem_fn_sig<Sig>:: template mem_fn_inst<inst>:: template base<Base>
{};

很明显，这个版本比前一个版本更多的代码。但好处是使用该工具的语法简单直观。事实上，语法类似于您原来的设施。我只是添加了一些代码来简化 MSVC 的编译过程。

通常，您会像这样使用该工具：

... = c_interface_gen<decltype(&Derived::f), &Derived::f, Base>::invoke;

如果Derived::f 被重载，您必须像这样显式指定其类型：

... = c_interface_gen<void(Derived::*)() const, &Derived::f, Base>::invoke;

注意，这里不需要为const成员函数指定const Base。您只需指定基本类型。模板会自动判断是否应该添加 const 修饰符。

以下是您使用第二个版本的示例代码：

#include <iostream>

template <typename Sig>
struct mem_fn_sig;

template <typename R, typename D, typename... Args>
struct mem_fn_sig<R(D::*)(Args...)> {
  template <R(D::*mem_fn)(Args...)>
  struct mem_fn_inst {
    template <typename Base>
    struct base {
      inline static R invoke(Base* pb, Args... args) {
        return (static_cast<D*>(pb)->*mem_fn)(args...);
      }
    };
  };
};

template <typename R, typename D, typename... Args>
struct mem_fn_sig<R(D::*)(Args...) const> {
  template <R(D::*mem_fn)(Args...) const>
  struct mem_fn_inst {
    template <typename Base>
    struct base {
      inline static R invoke(const Base* pb, Args... args) {
        return (static_cast<const D*>(pb)->*mem_fn)(args...);
      }
    };
  };
};

template <typename Sig, Sig inst, typename Base>
struct c_interface_gen:
  mem_fn_sig<Sig>:: template mem_fn_inst<inst>:: template base<Base>
{};

//
// C interface and function(s) typically defined elsewhere
//
#ifdef __cplusplus
extern "C" {
#endif

  // The C interface implemented by a 'greeter'
  struct greeter_c {
    void(*greet_cb)(const struct greeter_c * greeter,
                    const char * recipient);
  };

  // Some C function that makes use of a greeter
  void broadcast(const struct greeter_c * greeter) {
    greeter->greet_cb(greeter, "world");
  }

#ifdef __cplusplus
}  // extern "C"
#endif

//
// A C++ class that implements the C 'greeter' interface
//
class Greeter : public greeter_c {
 public:
  // Constructor
  Greeter(const char * greeting) : m_greeting(greeting) {
    // Set up C interface callback by wrapping member function

    // !! The following line causes the Visual Studio compilation error !!
    greet_cb = c_interface_gen<decltype(&Greeter::greet), &Greeter::greet, greeter_c>::invoke;
  }

  // C++ member function that 'does' the greeting
  void greet(const char * recipient) const {
    std::cout << m_greeting << " " << recipient << std::endl;
  }

 private:
  const char * m_greeting;
};


// An application that greets using a Greeter's C interface
int main(int argc, char * argv[]) {
  // Create C++ object that implements C interface
  Greeter a("Hello");

  // Greet using Greeter's C interface
  broadcast(static_cast<const greeter_c *>(&a));

  return 0;
}