Question

我有一段代码，简化了一下，相当于以下编译和正常工作。

template <typename Interface, typename... Args>
struct factory_function {
  typedef function<shared_ptr<Interface> (Args...)> type;
};

template <typename Interface, typename Implementer, typename... Args>
shared_ptr<Interface> create_function(Args... args) {
  return make_shared<Implementer>(args...);
}

template <typename Interface, typename... Args>
  int register_factory(identifier id, typename factory_function<Interface, Args...>::type factory) {
}

int main(int argc, char *argv[]) {
  register_factory<Iface>(1000, create_function<Iface, Impl>);
  return 0;
}

但是当尝试在这样的结构中使用较新的using ... =构造而不是typedef时：

template <typename Interface, typename... Args>
using factory_function = function<shared_ptr<Interface> (Args...)>;

然后将typename factory_function<Interface, Args...>::type更改为factory_function<Interface, Args...>，我收到编译错误：

foo.cc: In function ‘int main(int, char**)’:
foo.cc:31:61: error: no matching function for call to ‘register_factory(int, <unresolved overloaded function type>)’
   register_factory<Iface>(1000, create_function<Iface, Impl>);
                                                         ^
foo.cc:31:61: note: candidate is:
foo.cc:17:5: note: template<class Interface, class ... Args> int register_factory(identifier, factory_function<Interface, Args ...>)
 int register_factory(identifier id, factory_function<Interface, Args...> factory) {
     ^
foo.cc:17:5: note:   template argument deduction/substitution failed:
foo.cc:31:61: note:   mismatched types ‘std::function<std::shared_ptr<Iface>(Args ...)>’ and ‘std::shared_ptr<Iface> (*)()’
   register_factory<Iface>(1000, create_function<Iface, Impl>);
                                                         ^
foo.cc:31:61: note:   could not resolve address from overloaded function ‘create_function<Iface, Impl>’

更新这是完整的，可编译的测试用例，使用g++ -std=c++11 foo.cc编译：

#include <functional>
#include <memory>

using namespace std;

typedef int identifier;

template <typename Interface, typename... Args>
struct factory_function {
  typedef function<shared_ptr<Interface> (Args...)> type;
};
//template <typename Interface, typename... Args>
//using factory_function = function<shared_ptr<Interface> (Args...)>;

template <typename Interface, typename Implementer, typename... Args>
shared_ptr<Interface> create_function(Args... args) {
  return make_shared<Implementer>(args...);
}

template <typename Interface, typename... Args>
int register_factory(identifier id, typename factory_function<Interface, Args...>::type factory) {
//int register_factory(identifier id, factory_function<Interface, Args...> factory) {
}

class Iface {
public:
  virtual void foo() = 0;
};

class Impl : public Iface {
public:
  virtual void foo() {}
};

int main(int argc, char *argv[]) {
  register_factory<Iface>(1000, create_function<Iface, Impl>);
  return 0;
}

注释行显示不的工作原理。

Answer 1

使用int register_factory(identifier id, typename factory_function<Interface, Args...>::type factory)，编译器无法推断类型Interface和Args，因此对register_factory<Iface>(1000, create_function<Iface, Impl>);的调用明确int register_factory(identifier id, typename factory_function<Interface>::type);

使用替代定义（使用）int register_factory(identifier id, factory_function<Interface, Args...>，编译器必须尝试推导Args（Interface明确设置为Iface）但不幸的是，转换是需要和编译器失败（std::shared_ptr<Iface> (*)()不是任何function<std::shared_ptr<Iface>(Args...)>）

的完全匹配

解决方法是您的第二个解决方案是强制create_function<Iface, Impl>的类型为std::function。以下内容可能有所帮助：

template <typename R, typename...Args>
std::function<R(Args...)> make_function(R (*f)(Args...))
{
    return f;
}

后来：

register_factory<Iface>(1000, make_function(create_function<Iface, Impl>));

Answer 2

你可以通过结合这两种方法使其发挥作用;）

template <typename Interface, typename... Args>
struct fa_fu_helper {
  typedef function<shared_ptr<Interface> (Args...)> type;
};

template <typename Interface, typename... Args>
using factory_function = typename fa_fu_helper<Interface, Args...>::type;

typedef有效，＆＃39;使用=＆＃39;没有按＆＃39;吨

2 个答案: