typedef

问题 I'm trying to call explicit constructor/destructor with traits in templatized function. template <int i> struct Traits { }; template <> struct Traits<1> { typedef Foo type_t; }; template <> struct Traits<2> { typedef Bar type_t; }; template <class Traits> void DoSomething(void* p_in) { typename Traits::type_t* p = reinterpret_cast<typename Traits::type_t*>(p_in); // this works. new (p) typename Traits::type_t; // neither of following two does work. p->~typename Traits::type_t(); p->typename

问题 Why is that case incorrect (it's logical) template <typename T> struct Der: public Base { typedef int T; T val; }; , but that case is correct? struct Base { typedef int T; }; template <typename T> struct Der: public Base { T val; }; The Standard 14.6.1/7 says: In the definition of a class template or in the definition of a member of such a template that appears outside of the template definition, for each base class which does not depend on a template-parameter (14.6.2), if the name of the

问题 I'm aware of the fact that the 'dependent names' are not visible to the compiler by default. But I was told in answers to other SO questions (here, here, and ultimately on the C++ faq) that a using declaration may help. So I tried. A template base class: // regardless of the fact that members are exposed... template<typename T> struct TBase { typedef T MemberType; MemberType baseMember; MemberType baseFunction() { return MemberType(); } }; And a derived class, using the base's members:

问题 I'm aware of the fact that the 'dependent names' are not visible to the compiler by default. But I was told in answers to other SO questions (here, here, and ultimately on the C++ faq) that a using declaration may help. So I tried. A template base class: // regardless of the fact that members are exposed... template<typename T> struct TBase { typedef T MemberType; MemberType baseMember; MemberType baseFunction() { return MemberType(); } }; And a derived class, using the base's members:

来源： https://stackoverflow.com/questions/9403768/data-definition-has-no-type-or-storage-class

来源： https://stackoverflow.com/questions/9403768/data-definition-has-no-type-or-storage-class

来源： https://stackoverflow.com/questions/44990068/using-c-typedef-using-type-alias

来源： https://stackoverflow.com/questions/44990068/using-c-typedef-using-type-alias

来源： https://stackoverflow.com/questions/40332980/c-concepts-lite-and-type-alias-declaration