既指向成员函数的指针又指向const成员函数的指针的函数 [英] Function taking both pointer to member-function and pointer to const member-function
问题描述
我有以下代码库:
template <typename Type>
class SomeClass {
public:
template <typename ReturnType, typename... Params>
void register_function(const std::pair<std::string, ReturnType (Type::*)(Params...)> fct) {
auto f = [fct](Params... params) -> ReturnType { return (Type().*fct.second)(std::ref(params)...); }
// ...
}
};
当我将指针传递给成员函数(非const)时,此方法有效。
但是,如果我想传递一个指向const成员函数的指针,则会导致编译错误,我必须复制以上函数才能获得此代码:
This works when I pass a pointer to a member-function (non-const). However, if I want to pass a pointer to a const member-function, it results in a compile error and I must duplicate the above function to get this code:
template <typename Type>
class SomeClass {
public:
template <typename ReturnType, typename... Params>
void register_function(const std::pair<std::string, ReturnType (Type::*)(Params...)> fct) {
auto f = [fct](Params... params) -> ReturnType { return (Type().*fct.second)(std::ref(params)...); }
// ...
}
template <typename ReturnType, typename... Params>
void register_function(const std::pair<std::string, ReturnType (Type::*)(Params...) const> fct) {
auto f = [fct](Params... params) -> ReturnType { return (Type().*fct.second)(std::ref(params)...); }
// ...
}
};
现在,我既可以传递const成员函数,也可以传递非const成员函数。但是,现在,代码是重复的,并且降低了可维护性。
Now, I can pass both const-member-functions and non-const-member-functions. But, now, the code is duplicate and maintainability is reduced.
是否可以将这两个函数合并为一个既包含const成员函数又包含非成员函数的函数。 const-member-functions?
Is there a way to merge these two functions into a function taking both const-member-functions and non-const-member-functions?
重要说明:我必须真正将指针函数用作参数(无std :: function)。
Important note: I must really take a pointer function as parameter (no std::function).
编辑:我添加了更多代码。
在函数内部,我构建了一个与成员函数签名匹配的闭包(相同的返回类型和参数)。
此闭包将被存储并在以后用于反射(更多内容)
I've added a little bit more code. Inside the functions, I build a closure matching the member function signature (same return types and params). This closure will be stored and used later for making reflection (more here)
推荐答案
您可以编写类型特征,基于该特征可以告诉您如果某些 MF
是 Type
类型的成员函数指针:
You could write a type trait, based on which will tell you if some MF
is a pointer-to-member function on Type
:
template <typename C, typename T>
struct is_pointer_to_member_helper : std::false_type { };
template <typename C, typename T>
struct is_pointer_to_member_helper<C, T C::*> : std::is_function<T> { };
template <typename C, typename T>
struct is_pointer_to_member : is_pointer_to_member_helper<C,
std::remove_cv_t<T>
> { };
并使用它来确保只获得其中之一:
And use it to ensure that you only get one of those:
template <typename Type>
class SomeClass {
public:
template <typename MF>
std::enable_if_t<is_pointer_to_member<Type, MF>::value>
register_function(const std::pair<std::string, MF> fct)
{
auto f = [fct](auto&&... params) {
return (Type{}.*fct.second)(std::forward<decltype(params)>(params)...);
};
// ...
}
};
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