Actually, things got much easier in C++11 thanks to the decltype and late return bindings machinery. Now, it’s just simpler to use methods to test this: // Culled by SFINAE if reserve does not exist or is not accessible template <typename T> constexpr auto has_reserve_method(T& t) -> decltype(t.reserve(0), bool()) { return true; } // Used … Read more
Passing a pointer to base_all<U…> merely requires the existence of a declaration of base_all<U…>. Without attempting the to access the definition, the compiler won’t detect that the type is actually ill-defined. One approach to mitigate that problem would be to use an argument which requires a definition of base_all<U…>, e.g.: template< class …T> struct base_all … Read more
I think this trait does what you want. It detects operator() with any kind of signature even if it’s overloaded and also if it’s templatized: template<typename T> struct is_callable { private: typedef char(&yes)[1]; typedef char(&no)[2]; struct Fallback { void operator()(); }; struct Derived : T, Fallback { }; template<typename U, U> struct Check; template<typename> static … Read more
It returns true for both classes and structs. I know that in C++ they are almost the same thing, but I’d like to know why there’s not a distinction between them in the type trait. Unfortunately this is a common misconception in C++. Sometimes it comes from fundamental misunderstanding, but at other times it comes … Read more
If they are related Let’s for a moment assume that B is actually a base of D. Then for the call to check, both versions are viable because Host can be converted to D* and B*. It’s a user defined conversion sequence as described by 13.3.3.1.2 from Host<B, D> to D* and B* respectively. For … Read more