Lya*_*yar 6 c++ visual-c++ c++11
我正在努力在 C++11 中创建一个简单的反射器,它将实例函数的函数指针存储为:
static std::unordered_map<std::string, std::pair<void(EmptyClass::*)(void), int>>* methods;
template<typename ClassType, typename returnType, typename... Args>
static void RegistFunction(std::string name, returnType(ClassType::* func)(Args... args)) {
(*methods)[name] = std::make_pair((void(EmptyClass::*)())func, sizeof...(Args));
}
template<typename ReturnType, typename ClassType, typename... Args>
static ReturnType ExecuteFunction(ClassType* object, std::string name, Args... args) {
if (object == NULL) return;
ReturnType(ClassType:: * func)(Args...) = (ReturnType(ClassType::*)(Args...))(*methods)[name].first;
return (object->*func)(std::forward<Args>(args)...);
}
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但是当我想调用ExecuteFunction时,参数的数量可能比函数指针实际接受的数量多。所以我需要从参数列表的尾部删除一些参数,但似乎我只能从头部删除。
template<typename ReturnType, typename ClassType, typename Arg, typename... Args>
static ReturnType ExecuteFunction(ClassType* object, std::string name, Arg arg, Args... args) {
if (sizeof...(Args) + 1 > (*methods)[name].second) {
return ExecuteFunction<ReturnType>(std::forward<ClassType*>(object), std::forward<std::string>(name), std::forward<Args>(args)...);
}
if (object == NULL) return;
ReturnType(ClassType:: * func)( Arg, Args...) = (ReturnType(ClassType::*)(Arg, Args...))(*methods)[name].first;
return (object->*func)(std::forward<Arg>(arg), std::forward<Args>(args)...);
}
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有没有解决方案可以删除可变参数方法模板尾部的参数?
std::string这是仅依赖于和 的C++11 实现std::unordered_map。一些强制性备注:
method确实不应该是一个指针。和朋友的 C++11 实现可以在这里std::index_sequence找到。
template<typename...>
struct typelist {};
template<size_t, typename, typename, typename, typename>
struct call;
template<size_t N, typename R, typename C, typename... Accum, typename Head, typename... Tail>
struct call<N, R, C, typelist<Accum...>, typelist<Head, Tail...>>
: call<N, R, C, typelist<Accum..., Head>, typelist<Tail...>>
{
};
template<typename R, typename C, typename... Accum, typename Head, typename... Tail>
struct call<sizeof...(Accum), R, C, typelist<Accum...>, typelist<Head, Tail...>>
{
template<typename... Ts>
int operator()(Ts&&...)
{
return 0;
}
template<typename... Ts>
int operator()(R& ret, void (EmptyClass::* g)(), C& obj, Accum&... args, Ts&&...)
{
auto f = (R (C::*)(Accum...))g;
ret = (obj.*f)(std::move(args)...);
return 0;
}
};
template<typename R, typename C, typename... Args, size_t... Is>
R switcher(int i, index_sequence<Is...>, void (EmptyClass::* g)(), C& obj, Args&... args)
{
R ret{};
int unused[] = {(i == Is ?
call<Is, R, C, typelist<>, typelist<Args..., void>>{}(ret, g, obj, args...)
: 0)...};
(void)unused;
return ret;
}
template<typename C, typename R, typename... Args>
void reg(std::string name, R (C::* func)(Args... args)) {
(*methods)[name] = std::make_pair((void (EmptyClass::*)())func, sizeof...(Args));
}
template<typename R, typename C, typename... Args>
R exec(C* obj, std::string name, Args... args) {
if(obj == nullptr)
throw "a tantrum";
auto& info = (*methods)[name];
auto g = info.first;
size_t i = info.second;
if(i > sizeof...(Args))
throw "a fit";
return switcher<R>(i, make_index_sequence<sizeof...(Args) + 1>{}, g, *obj, args...);
}
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