I think it is referring to the following case: template <typename T> struct base { void foo() { std::cout << “generic” << std::endl; } void bar() { std::cout << “bar” << std::endl; } }; template <> void base<int>::foo() // specialize only one member { std::cout << “int” << std::endl; } int main() { base<int> i; … Read more
You cannot partially specialize only a single member function, you must partially specialize the whole class. Hence you’ll need something like: template <typename T> class Object<T, 0> { private: T m_t; Object(); public: Object(T t): m_t(t) {} T Get() { return m_t; } Object& Deform() { std::cout << “Spec\n”; m_t = -1; return *this; } … Read more
Though it is an old and outdated question, it may worth noting that C++11 had solved this issue using deleted functions: template<typename T> T GetGlobal(const char *name) = delete; template<> int GetGlobal<int>(const char *name); UPDATE This will not compile under MacOS llvm 8. It is due to a still hanging 4 years old defect (see … Read more
The default argument applies to the specialization — and, in fact, a specialization must accept (so to speak) the base template’s default argument(s). Attempting to specify a default in the specialization: template<class A = int, class B=double> class Base {}; template<class B=char> // … …is an error. Likewise, if we change the specialization so that … Read more
In C++11 and later, is it allowed to specialize std::to_string in the std namespace for custom types? No. First of all, it is not a template function so you can’t specialize it at all. If you’re asking about adding your own overload functions the answer still remains the same. Documentation snippet from Extending the namespace … Read more
It sounds like you want to change: friend ostream& operator << (ostream& out, const Base<T>& e); To: template<class T> friend ostream& operator << (ostream& out, const Base<T>& e);
The default argument applies to the specialization — and, in fact, a specialization must accept (so to speak) the base template’s default argument(s). Attempting to specify a default in the specialization: template<class A = int, class B=double> class Base {}; template<class B=char> // … …is an error. Likewise, if we change the specialization so that … Read more
It doesn’t work that way. You would need to say the following, but it is not correct template <class C> template<> void X<C>::get_as<double>() { } Explicitly specialized members need their surrounding class templates to be explicitly specialized as well. So you need to say the following, which would only specialize the member for X<int>. template … Read more
As with simple functions you can use declaration and implementation. Put in your header declaration: template <> void TClass<int>::doSomething(std::vector<int> * v); and put implementation into one of your cpp-files: template <> void TClass<int>::doSomething(std::vector<int> * v) { // Do somtehing with a vector of int’s } Don’t forget to remove inline (I forgot and thought this … Read more