我已经实现了这样的元编程成员检查器:
//Declare a metaprogramming test to check if a class has a member
//of a given signature and a certain name
#define DECL_hasMember(MEMBER) \
/* Check if T has member MEMBER M is a pointer to member of T type (T::*) */ \
template <typename T, typename M> \
class hasMember_##MEMBER { \
private: \
typedef char passed[1]; \
typedef char failed[2]; \
\
template <typename U, U> struct reallyHas; \
\
template <typename C> static passed& test(reallyHas<M, &C::MEMBER>*); \
template <typename C> static failed& test(...); \
\
hasMember_##MEMBER()=delete; \
public: \
static bool const value = sizeof(test<T>(nullptr)) == sizeof(passed); \
}
我已经在很多情况下测试过了,检查器在大多数情况下都可以正常工作,
现在我正在使用奇怪的重复模板模式实现有限状态机,它有一个将状态关联到派生类函数的映射。我希望这个派生类有一个名为 buildFSM 的成员,用于创建所述 map 。我想使用静态断言来强制执行此操作,以获得更好的编译器错误。但是元编程检查器似乎无法检测到 desc_t
(为了这个问题,我试图在一个更简单的场景中复制这种情况,但无法让它在不同的上下文中失败)
DECL_hasMember(buildFSM);
template<class Executor_T, typename State_T = uint32_t, typename ... Args>
class FSM
{
public:
typedef State_T state_t;
typedef FSM<Executor_T, State_T, Args...> fsm_t;
protected:
/* returns the next state */
typedef state_t (Executor_T::*handler_t)(Args...);
/* holds all the states */
struct desc_t : private std::map<state_t, handler_t> {
friend fsm_t;
void addState(const state_t&, handler_t);
using std::map<state_t, handler_t>::size;
using std::map<state_t, handler_t>::empty;
};
private:
/* Implementation must implement buildFSM returning a desc_t that initializes this */
static desc_t states;
static_assert(hasMember_buildFSM<Executor_T, desc_t(Executor_T::*)()>::value,
"Implementing class must provide desc_t buildFSM() function member.");
/* [...] */
};
//Delegate the initialization of states to the derived class
template<class E, typename S, typename ... A>
decltype(FSM<E, S, A...>::states) FSM<E, S, A...>::states= std::move(E::buildFSM());
struct test : public FSM<test> {
fsm_t::desc_t buildFSM() {
return fsm_t::desc_t();
}
};
正如我之前所说,静态断言失败,即它无法检测到 test::buildFSM
。问题出在 desc_t
中,因为如果我将其更改为另一种类型,测试就会顺利通过。如果我修改 desc_t
使其不派生 std::map
,测试也会通过。为什么会这样?在这种情况下,我能做些什么来完成这项工作吗?我知道另一种方法是将 map 作为成员而不是基类,但我不明白为什么会这样。
我在 Visual Studio 2015 中。
最佳答案
下面是一些检测成员函数的代码:
#include <iostream>
// single argument and return type
#define DECL_hasMemberRetArg(MEMBER) \
template<typename T, typename RESULT, typename ARG1> \
class Detect_RetArg_##MEMBER \
{ \
template <typename U, RESULT (U::*)(ARG1)> struct Check; \
template <typename U> static char func(Check<U, &U::MEMBER> *); \
template <typename U> static int func(...); \
public: \
typedef Detect_RetArg_##MEMBER type; \
enum { value = sizeof(func<T>(0)) == sizeof(char) }; \
}
// two arguments and return type
#define DECL_hasMemberRetArgArg(MEMBER) \
template<typename T, typename RESULT, typename ARG1, typename ARG2> \
class Detect_RetArgArg_##MEMBER \
{ \
template <typename U, RESULT (U::*)(ARG1, ARG2)> struct Check; \
template <typename U> static char func(Check<U, &U::MEMBER> *); \
template <typename U> static int func(...); \
public: \
typedef Detect_RetArgArg_##MEMBER type; \
enum { value = sizeof(func<T>(0)) == sizeof(char) }; \
}
class a_t
{
char get(int);
char put(int,long);
};
struct b_t
{
char get(int);
char put(int,long);
};
struct c_t : public b_t
{
};
DECL_hasMemberRetArg(get);
DECL_hasMemberRetArgArg(put);
int main()
{
std::cout << Detect_RetArg_get<a_t, char, int>::value << "\n"; // 0
std::cout << Detect_RetArgArg_put<a_t, char, int, long>::value << "\n"; // 0
std::cout << Detect_RetArg_get<b_t, char, int>::value << "\n"; // 1
std::cout << Detect_RetArgArg_put<b_t, char, int, long>::value << "\n"; // 1
std::cout << Detect_RetArg_get<c_t, char, int>::value << "\n"; // 0
std::cout << Detect_RetArgArg_put<c_t, char, int, long>::value << "\n"; // 0
return 0;
}
模板看不到基类成员。
关于c++ - 元编程成员检查器不适用于返回具有基类的类型的函数成员,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33660319/