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template specialization compiler error

user7460 Published in September 21, 2018, 8:36 am

I have a main.cpp and a header file called bar.h This is the main.cpp. My goal is to have the program print out "bar":

#include "bar.h"

struct Raw{

template<typename Obj>
struct IsBar{
    static const bool value = false;

template<typename Obj, bool>
struct function{
    static void callbar(Obj obj){ 

template<typename Obj>
struct function<Obj, false>{
    static void callbar(Obj obj){ 
        std::cout<< "no bar()" << std::endl;

int main()
    typedef Bar<Raw> Obj;
    Obj obj;

    function<Obj, IsBar<Obj>::value> f;

    return 0;

For the bar.h:

template<typename T>
struct Bar{
    void bar()
        std::cout<< "bar" << std::endl;

struct IsBar<Bar>{ // I know this wouldn't work, but how do I do something like Bar<Raw> instead of just Bar?
    static const bool value = true;

Compiler gave me this error: error: 'IsBar' is not a class template. Previously I tried to have the content of bar.h inside the main.cpp, and everything works fine, because Raw is known when I declared the IsBar<> specialization using Bar<Raw>.

share|improve this question
  • The template must be defined before the specialization of it. – immibis Feb 12 at 2:45
  • Of course you can do template<> struct IsBar<Bar<Raw>>, or even template<typename T> struct IsBar<Bar<T>>, but the primary template must be visible when you add the specialization. – Bo Persson Feb 12 at 2:45

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