Override template member in Interface

Question

Is it possible to declare some type of base class with template methods which i can override in derived classes? Following example:

#include          


        

Answer 1

I had the same problem, but I actually came up with a working solution. The best way to show the solution is by an example:

What we want(doesn't work, since you can't have virtual templates):

class Base
{
    template <class T>
    virtual T func(T a, T b) {};
}

class Derived
{
    template <class T>
    T func(T a, T b) { return a + b; };
}

int main()
{
    Base* obj = new Derived();
    std::cout << obj->func(1, 2) << obj->func(std::string("Hello"), std::string("World")) << obj->func(0.2, 0.1);
    return 0;
}

The solution(prints 3HelloWorld0.3):

class BaseType
{
public:
    virtual BaseType* add(BaseType* b) { return {}; };
};

template <class T>
class Type : public BaseType
{
public:
    Type(T t) : value(t) {};
    BaseType* add(BaseType* b)
    {
        Type<T>* a = new Type<T>(value + ((Type<T>*)b)->value);
        return a;
    };

    T getValue() { return value; };
private:
    T value;
};

class Base
{
public:
    virtual BaseType* function(BaseType* a, BaseType* b) { return {}; };
    template <class T>
    T func(T a, T b)
    {
        BaseType* argA = new Type<T>(a);
        BaseType* argB = new Type<T>(b);
        BaseType* value = this->function(argA, argB);
        T result = ((Type<T>*)value)->getValue();
        delete argA;
        delete argB;
        delete value;
        return result;
    };
};

class Derived : public Base
{
public:
    BaseType* function(BaseType* a, BaseType* b)
    {
        return a->add(b);
    };
};

int main()
{
    Base* obj = new Derived();
    std::cout << obj->func(1, 2) << obj->func(std::string("Hello"), std::string("World")) << obj->func(0.2, 0.1);
    return 0;
}

We use the BaseType class to represent any datatype or class you would usually use in a template. The members(and possibly operators) you would use in a template are described here with the virtual tag. Note that the pointers are necessary in order to get the polymorphism to work.

Type is a template class that extends Derived. This actually represents a specific type, for example Type<int>. This class is very important, since it allows us to convert any type into the BaseType. The definition of the members we described described in BaseType are implemented here.

function is the function we want to override. Instead of using a real template we use pointers to BaseType to represent a typename. The actual template function is in the Base class defined as func. It basically just calls function and converts T to Type<T>. If we now extend from Base and override function, the new overridden function gets called for the derived class.

Answer 2

Another possible aproach to make your example work as you expect is to use std::function:

class Base {
  public:
    Base() {
      virtualFunction = [] () -> string { return {"Base"}; };
    }
    template <class T> string do_smth() { return virtualFunction(); }
    function<string()> virtualFunction;
};
class Derived : public Base {
  public:
    Derived() {
      virtualFunction = [] () -> string { return {"Derived"}; };
    }
};

int main() {
  auto ptr = unique_ptr<Base>(new Derived);
  cout << ptr->do_smth<bool>() << endl;
}

This outputs "Derived". I'm not sure that this is what you realy want, but I hope it will help you..

Answer 3

1) Your functions, in order to be polymorphic, should be marked with virtual

2) Templated functions are instantiated at the POI and can't be virtual (what is the signature??How many vtable entries do you reserve?). Templated functions are a compile-time mechanism, virtual functions a runtime one.

Some possible solutions involve:

• Change design (recommended)
• Follow another approach e.g. multimethod by Andrei Alexandrescu (http://www.icodeguru.com/CPP/ModernCppDesign/0201704315_ch11.html)

Answer 4

Template methods cannot be virtual. One solution is to use static polymorphism to simulate the behavior of "template virtual" methods:

#include <iostream>
#include <stdexcept>
#include <string>

template<typename D>
class Base
{
    template<typename T>
    std::string _method() { return "Base"; }
public:

    template<typename T>
    std::string method()
    {
       return static_cast<D&>(*this).template _method<T>();
    }
};

class Derived : public Base<Derived>
{
    friend class Base<Derived>;

    template<typename T>
    std::string _method() { return "Derived"; }
public:
    //...
};

int main()
{
    Base<Derived> *b = new Derived();
    std::cout << b->method<bool>() << std::endl;
    return 0;
}

where method is the interface and _method is the implementation. To simulate a pure virtual method, _method would absent from Base.

Unfortunately, this way Base changes to Base<Derived> so you can no longer e.g. have a container of Base*.

Also note that for a const method, static_cast<D&> changes to static_cast<const D&>. Similarly, for an rvalue-reference (&&) method, it changes to static_cast<D&&>.