How to derive a non-template class from a template base class

Question

I've a simple problem but I really can't understand why. I think you can help me.

I have a template base class :

template <class T> class Base
{
    public:
        T foo(T t);
    protected:

        Base();
        ~Base();
};

I want to derive a specialisation of this template :

class Derived : public Base<std::string>
{
    public:
        Derived();
        virtual ~Derived();
};

All the functions are defined in the respective .cpp (they are empty, except for foo which just returns t)

I chose std::string as an example. But this simple code doesn't compile. I have errors : "indefinite reference to "Base::Base()" "indefinite reference to "Base::~Base()"

It seems that I follow examples on internet but it doesn't work... I really don't understand this one, it must be obvious!

Can you help me? :)

(PS : sorry if my english is bad)

Not sure if it's *the* problem, but `Base::~Base()` should be virtual. — Biffen, Apr 11 '16 at 09:40
It's not useful to describe what the errors are *like*. Tell us *exactly* what the errors are. — eerorika, Apr 11 '16 at 09:44
@matthieuperez Well, it still should. It's hard to say what the problem is without seeing more code. Could you post an [MCVE](http://stackoverflow.com/help/mcve), including the *exact* errors? — Biffen, Apr 11 '16 at 09:44
Do you provide definitions for `Base()`, `~Base()`, `Derived()`, `~Derived()` anywhere ? Are the definitions for `Base()` and `~Base()` in the same translation unit as the corresponding declaration of class `B` ? — Piotr Skotnicki, Apr 11 '16 at 09:45
Your program compiles just fine, without errors: http://coliru.stacked-crooked.com/a/8f7a8dc11e1d09a5 You need to create a [mcve] — eerorika, Apr 11 '16 at 09:46
Do you provide definitions of `Base` methods into a cpp ? if so, it should be in header instead. — Jarod42, Apr 11 '16 at 09:50
voted to close. as unclear. there is no error in the code in the post. — Richard Hodges, Apr 11 '16 at 09:54
It works when I define base functions inside the header file, instead of inside the .cpp... Thank you ! But... why? — matthieu perez, Apr 11 '16 at 09:58
You cannot just separate headers and implementation for templated classes. There are hundreds of questions on that on SO. — Chiel, Apr 11 '16 at 10:02
Oh, ok :) I didn't know that. I'll search. Thank you for your help ! — matthieu perez, Apr 11 '16 at 10:05

aybassiouny · Accepted Answer · 2016-04-11T10:32:18.643

When you inherit from a class template, base class declaration and definition need to be in the header file.

The reason for this is that compiler needs to instantiate the template in compile time for the types you use, during compile time. Foreign Cpp files are only accessed during linkage, hence when compiler tries to instantiate derived class it cannot instantiate template code and the error you mentioned results.

score 0 · Answer 2 · answered Apr 11 '16 at 10:00

0

This example compiles perfectly, so where is your problem?

#include <string>

template <class T> class Base
{
    public:
        T foo(T t) {};
    protected:
        Base() {};
        ~Base() {};
};

class Derived : public Base<std::string>
{
    public:
        Derived() {};
        virtual ~Derived() {};
};

int main()
{
    Derived derived;
    return 0;
}

answered Apr 11 '16 at 10:00

Chiel

6,006
2
32
57

It seems that it is necessary to define the function inside the .h, and not the .cpp... Could you tell me why? – matthieu perez Apr 11 '16 at 10:03
See my other comment. If you use templates you have to put your implementation inside of the header, or you have to resort to other tricks. See http://stackoverflow.com/questions/1724036/splitting-templated-c-classes-into-hpp-cpp-files-is-it-possible – Chiel Apr 11 '16 at 10:04
It may compile but any function that calls `Base::foo()` and uses its return value will have undefined behaviour. – Peter Apr 11 '16 at 10:18

How to derive a non-template class from a template base class

2 Answers2