英文:
std::is_member_function_pointer not working for overloaded functions?
问题
我已经编写了一个特性，通过std::is_member_function_pointer来检测一个类是否具有特定的公共成员函数：
然而，似乎当函数被重载时，std::is_member_function_pointer无法检测到它，如Test2所示。是否有办法使它能够在方法被重载时正常工作？
英文:
I've written a trait that detects whether a class has a certain public member function via std::is_member_function_pointer:
class Test1{public:void method();};class Test2{public:void method();void method(int);};template <typename T, typename = void>struct HasMethod : std::false_type{};template <typename T>struct HasMethod<T, std::enable_if_t<std::is_member_function_pointer<decltype(&T::method)>::value>>: std::true_type{};int main(){std::cout << HasMethod<Test1>::value << std::endl; // prints 1std::cout << HasMethod<Test2>::value << std::endl; // prints 0}
However it seems that when the function is overloaded, then std::is_member_function_pointer is not able to detect it, as shown with Test2.
Is there a way to make it work regardless of whether the method is overloaded?
答案1
得分: 1
以下是代码部分的翻译：
要检测类成员的存在通常的方法是使用`std::void_t`模板。您可以像这样执行它：```c++#include <iostream>#include <type_traits>class Test1{public:void method();};class Test2{public:void method();void method(int);};template <typename T, typename = void>struct HasMethod : std::false_type{};// 使用method(0)来检测void method(int)的重载// 在这种情况下，您甚至不需要std::void_t，因为method无论如何返回void，但对于任何其他返回类型，您都需要它template <typename T>struct HasMethod<T, std::void_t<decltype(std::declval<T>().method())>>: std::true_type{};int main(){std::cout << HasMethod<Test1>::value << std::endl; // 输出1std::cout << HasMethod<Test2>::value << std::endl; // 输出1}
```c++好的。经过一些思考，我找到了解决方案。我们可以利用`declval(&T::method)`将失败的事实，如果有多于一个重载，通过添加另一个来检测我们是否至少有一个重载。以下是解决方案。它相当冗长，但我无法缩减它。至少它可以工作。```c++#include <iostream>#include <type_traits>class Test{};class Test1{public:void method();};class Test2{public:void method();void method(int);};class Test3{public:using method = int;};class Test4{public:int method;};template<typename T, typename = void>struct HasSingleOverload : std::false_type {};template<typename T>struct HasSingleOverload<T, std::void_t<decltype(&T::method)>> : std::true_type {};template<typename T, typename = void>struct IsMemberFunction : std::false_type {};template<typename T>struct IsMemberFunction<T, std::enable_if_t<std::is_member_function_pointer<decltype(&T::method)>::value>> : std::true_type {};template<typename T, typename = void>struct IsType : std::false_type {};template<typename T>struct IsType<T, std::void_t<typename T::method>> : std::true_type {};struct HasOverload {void method();};template<typename T>struct CheckOverload : T, HasOverload {};template<typename T>using HasConflict = std::bool_constant<!HasSingleOverload<CheckOverload<T>>::value>;template<typename T>using HasAnyOverload = std::conditional_t<HasSingleOverload<T>::value || IsType<T>::value, IsMemberFunction<T>, HasConflict<T>>;int main(){std::cout << HasAnyOverload<Test>::value << std::endl; // 输出0std::cout << HasAnyOverload<Test1>::value << std::endl; // 输出1std::cout << HasAnyOverload<Test2>::value << std::endl; // 输出1std::cout << HasAnyOverload<Test3>::value << std::endl; // 输出0std::cout << HasAnyOverload<Test4>::value << std::endl; // 输出0}
这是您提供的代码的翻译。<details><summary>英文:</summary>To detect existence of class members usual approach is to use `std::void_t` template. You can do it like this:```c++#include <iostream>#include <type_traits>class Test1{public:void method();};class Test2{public:void method();void method(int);};template <typename T, typename = void>struct HasMethod : std::false_type{};// Use method(0) to detect void method(int) overload// You don't even need std::void_t in this case, since method returns void anyway, but for any other return type you will need ittemplate <typename T>struct HasMethod<T, std::void_t<decltype(std::declval<T>().method())>>: std::true_type{};int main(){std::cout << HasMethod<Test1>::value << std::endl; // prints 1std::cout << HasMethod<Test2>::value << std::endl; // prints 1}
Ok. After some thinking I found solution. We can use fact that declval(&T::method) will fail, if there is more than one overload, to detect if we have at least one overload, by adding another one. Here is solution. It is quite verbose, but I was unable to reduce it. At least it works.
#include <iostream>#include <type_traits>class Test{};class Test1{public:void method();};class Test2{public:void method();void method(int);};class Test3{public:using method = int;};class Test4{public:int method;};template<typename T, typename = void>struct HasSingleOverload : std::false_type {};template<typename T>struct HasSingleOverload<T, std::void_t<decltype(&T::method)>> : std::true_type {};template<typename T, typename = void>struct IsMemberFunction : std::false_type {};template<typename T>struct IsMemberFunction<T, std::enable_if_t<std::is_member_function_pointer<decltype(&T::method)>::value>> : std::true_type {};template<typename T, typename = void>struct IsType : std::false_type {};template<typename T>struct IsType<T, std::void_t<typename T::method>> : std::true_type {};struct HasOverload {void method();};template<typename T>struct CheckOverload : T, HasOverload {};template<typename T>using HasConflict = std::bool_constant<!HasSingleOverload<CheckOverload<T>>::value>;template<typename T>using HasAnyOverload = std::conditional_t<HasSingleOverload<T>::value || IsType<T>::value, IsMemberFunction<T>, HasConflict<T>>;int main(){std::cout << HasAnyOverload<Test>::value << std::endl; // prints 0std::cout << HasAnyOverload<Test1>::value << std::endl; // prints 1std::cout << HasAnyOverload<Test2>::value << std::endl; // prints 1std::cout << HasAnyOverload<Test3>::value << std::endl; // prints 0std::cout << HasAnyOverload<Test4>::value << std::endl; // prints 0}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论