[Thomas Klimpel]
Find attached a test case showing the scenario I described. This code actually really triggers the mentioned compile error. Standard wording can't help us here, because the standard doesn't say anything about dynamic link libraries. So I wouldn't see this as a bug, even if it is annoying at times.
By design. C:\Temp>type meow.cpp template <class T> struct container { T m; void forward_member_function() { m.member_function(); } }; struct b { }; #ifdef BOOM struct __declspec(dllexport) my_class : public container<b> { }; #else struct my_class : public container<b> { }; #endif C:\Temp>cl /EHsc /nologo /W4 /c meow.cpp meow.cpp C:\Temp>cl /EHsc /nologo /W4 /c /DBOOM meow.cpp meow.cpp meow.cpp(5) : error C2039: 'member_function' : is not a member of 'b' meow.cpp(9) : see declaration of 'b' meow.cpp(4) : while compiling class template member function 'void container<T>::forward_member_function(void)' with [ T=b ] meow.cpp(12) : see reference to class template instantiation 'container<T>' being compiled with [ T=b ] http://msdn.microsoft.com/en-us/library/81h27t8c.aspx says: "Inheritance and Exportable Classes All base classes of an exportable class must be exportable." http://msdn.microsoft.com/en-us/library/twa2aw10.aspx says: "the compiler changed the semantics of dllexport when it is applied to a class that has one or more base-classes and when one or more of the base classes is a specialization of a class template. In this case, the compiler implicitly applies dllexport to the specializations of class templates." (This page is a little confusing because it was written when VS 2003 was new.) When my_class is dllexported, its base class container<b> is explicitly instantiated and compiled into the DLL. That's why container<b>::forward_member_function() is being instantiated and subsequently exploding. I'm not an expert when it comes to DLLs, but I can't imagine another way for this to physically work. In the absence of dllexport, the compiler plays by the Standard rules that allow std::list<T> to work when T doesn't have op<(), as long as std::list<T>::sort() isn't called. Stephan T. Lavavej Visual C++ Libraries Developer