c++ - Is there any way round to make this template specialisation link? -

i have tried following hierarchy , doesn't link. call c->execute() not seen since seems masked execute in derived , doesn't find adequate type use. have thought if there problem, have show @ compilation. have tried gcc 4.8 online , borland compiler on windows. error message similar in both cases. in order of:

/tmp/ccrt5mny.o:(.rodata._ztv7derivedi5typece[_ztv7derivedi5typece]+0x10): undefined reference `derived<typec>::execute()' collect2: error: ld returned 1 exit status 

i grateful pointers.

#include <iostream>  class basetype  { public:      basetype() {}     virtual void execute() { std::cout << "basetype execute..." << std::endl; };  protected: };  struct typea; struct typeb; struct typec;  class derived2 : public basetype { public:     derived2() : basetype() {}     virtual void execute() { std::cout << "derived execute2" << std::endl; }  protected: };   template <typename t> class derived : public basetype { public:     derived() : basetype() {}     virtual void execute();  protected: };  template <typename t> class grandchild : public derived<t> { public:     grandchild() : derived<t>() {}     void execute(); };  template<> void derived<typea>::execute() { std::cout << "derived execute... typea" << std::endl; } template<> void derived<typeb>::execute() { std::cout << "derived execute... typeb" << std::endl; } template<> void grandchild<typec>::execute() { std::cout << "derived execute... typec" << std::endl; }   int main() {     basetype* = new derived<typea>();           basetype* b = new derived<typeb>();           basetype* c = new grandchild<typec>();           basetype* d2 = new derived2();     a->execute();     b->execute();     c->execute();     d2->execute();      delete a;     delete b;     delete c;     delete d2; } 

derived<typec>::execute() must defined if never call it, since you're instantiating class template. try adding empty definition, or 1 throws, etc. , program should link , work expect.