template-specialization

I have a template class with both a type and a non-type template parameter. I want to specialize a member function, what I finding is, as in the example below, I can do a full specialization fine. template<typename T, int R> struct foo { foo(const T& v) : value_(v) {} void bar() { std::cout << "Generic" << std::endl; for (int i = 0; i < R; ++i) std::cout << value_ << std::endl; } T value_; }; template<> void foo<float, 3>::bar() { std::cout << "Float" << std::endl; for (int i = 0; i < 3; ++i) std::cout << value_ << std::endl; } However this partial specialization won't compile. template<int R>

My third question here today ;-), but I am really new to c++ template programming and operator overloading. I am trying the following: terminallog.hh //snipped code class Terminallog { public: Terminallog(); Terminallog(int); virtual ~Terminallog(); template <class T> Terminallog & operator<<(const T &v); template <class T> Terminallog & operator<<(const std::vector<T> &v); template <class T> Terminallog & operator<<(const T v[]); Terminallog & operator<<(const char v[]); //snipped code }; //snipped code template <class T> Terminallog &Terminallog::operator<<(const T &v) { std::cout << std:

I can define a specialized function in a cpp like so... // header template<typename T> void func(T){} template<> void func<int>(int); // cpp template<> void func<int>(int) {} How can I define a method in a specialized class in a cpp? Like so (which doesn't work, I get error C2910: 'A<int>::func' : cannot be explicitly specialized )... // header template<typename T> struct A { static void func(T){} }; template<> struct A<int> { static void func(int); }; // cpp template<> void A<int>::func(int) {} Use following syntax in your .cpp file: void A<int>::func(int) { } This is Visual C++ kinda feature

Is there a better way to do the following? #include <iostream> template <typename T> T Bar(); template <> int Bar<int>() { return 3; } // Potentially other specialisations int main() { std::cout << Bar<int>() << std::endl; // This should work std::cout << Bar<float>() << std::endl; // This should fail } The problem with this solution is that it fails at (understandably) link time with "undefined reference to float Bar<float>() " or the like. This can be confusing for other developers as they may suspect an implementation file is not being linked. I do know another potential solution: template

template <typename Foo, Foo Part> struct TSelect {}; enum What { The }; template <typename Foo> struct AnotherOneSelector { static constexpr Foo Id = Foo::The; }; template <typename Foo, typename SelectPartType> struct THelper; template <typename Foo> struct THelper<Foo, TSelect<Foo, AnotherOneSelector<Foo>::Id>> {}; template <typename Foo, Foo PartId> struct THelper<Foo, TSelect<Foo, PartId>> {}; int main() { THelper<What, TSelect<What, What::The>> t; } This code compiles with gcc8.1 with each of the standard option (c++11, c++14, c++17), but clang trunk does not with c++17 (although with c+

##A.hh template<class T> void func(T t) {} template<> void func<int>(int t) {} void func2(); ##A.cpp void func2() {} ##main.cpp func("hello"); func(int()); The error I get is: error LNK2005: "void __cdecl func(int)" (??$func@H@@YAXH@Z) already defined in A.obj, one or more multiply defined symbols found Is a function template specialization not treated as a normal function template? It looks like it will be in the objective file for A. As template<> void func<int>(int t) {} is a function overload rather than a function template (i.e., all types are known at the point of definition so it is no

Is it possible to specialize an Iterator template parameter by its value_type ? I have a function with the following prototype. template<typename InputIterator> void f(InputIterator first, InputIterator last); And I want to handle specially if InputIterator::value_type is SomeSpecificType. Michael Anderson You can use some intermediate structs to get the partial template specialisation that you need. Something like this should do the trick template<typename T, typename V> struct f_impl { static void f( T first, T last ) {...}; //Default version }; template<typename T> struct f_impl<T,

There are cases where one uses an always_false helper to e.g. cause unconditional static_assert failure if instantiation of some template is attempted: template <class... T> struct always_false : std::false_type {}; template<class T> struct UsingThisShouldBeAnError { static_assert(always_false<T>::value, "You should not use this!"); }; This helper is necessary because a template definition must (at least theoretically) have at least one set of template parameters for which a valid specialization can be produced in order for the program to be well-formed: [temp.res]/8 : The program is ill