Downcasting from base pointer to templated derived types

Question

I have the following hierarchy:

class base
{
public:
   virtual ~base(){}
   virtual void foo() {}
};

template 
class derived1 : public base
{         


        

Answer 1

Solution 1: add one more virtual function:

enum DerivedType
{
    One,
    Two,
    ...
};

class base 
{ 
public: 
   ~base(){} 
   virtual void foo() {}
   virtual DerivedType GetType() = 0;
}; 

template <typename T> 
class derived1 : public base 
{ 
   virtual void foo() {};
   virtual DerivedType GetType() { return One; }
}; 

template <typename T> 
class derived2 : public base 
{ 
   virtual void foo() {};
    virtual DerivedType GetType() { return Two; }
}; 

Solution 2: using tag classes:

class Base
{
public:
    virtual ~Base() { }
};

class Derived1Tag
{ };

class Derived2Tag
{ };

template <class T>
class Derived1 : public Base, public Derived1Tag
{ };

template <class T>
class Derived2 : public Base, public Derived2Tag
{ };


int main(int argc, char** argv)
{
    Derived1<int> d1;
    Derived2<int> d2;

    cout << dynamic_cast<Derived1Tag*>((Base*)&d1) << endl;
    cout << dynamic_cast<Derived1Tag*>((Base*)&d2) << endl;

    return 0;
}

Answer 2

Move the logic which depends on the type into the type.

Instead of:

if (dynamic_cast<derived1<int>*>(b) ||
    dynamic_cast<derived1<unsigned int>*>(b) ||
    dynamic_cast<derived1<double>*>(b))
  std::cout << "is derived1";
else if (dynamic_cast<derived2<int>*>(b) ||
    dynamic_cast<derived2<unsigned int>*>(b) ||
    dynamic_cast<derived2<double>*>(b))
  std::cout << "is derived2";

add a virtual print_name() const function to base, and then do:

void example() {
    std::unique_ptr<base> b(new derived1<int>());
    b->print_name();
}
class base
{
public:
   ~base(){}
   virtual void foo() {}
   virtual void print_name() const = 0;
};

template <typename T>
class derived1 : public base
{
   virtual void foo() {}
   virtual void print_name() const {
       std::cout << "is derived1";
   }
};

template <typename T>
class derived2 : public base
{
   virtual void foo() {}
   virtual void print_name() const {
       std::cout << "is derived2";
   }
};

Answer 3

Insert a non-templated class inbetween base and derived1 or derived2:

class base
{
public:
   virtual ~base() {}  // **NOTE** Should be virtual
   virtual void foo() {}
};

class derived1_base : public base
{
};

template <typename T>
class derived1 : public derived1_base
{
public:
   virtual void foo() {}
};

class derived2_base : public base
{
};

template <typename T>
class derived2 : public derived2_base
{
public:
   virtual void foo() {}
};

In a comment, you mentioned:

[I want to] call a particular function for each one - btw there's more than derived1 and derived2

Add that (virtual) function to derived1_base, and you don't even need to know T anymore.

if (dynamic_cast<derived1_base*>(foo))
{
  std::cout << "is derived1";
  dynamic_cast<derived1_base*>(foo)->specific_derived1_function();
}
else if (dynamic_cast<derived2_base*>(foo))
{
  std::cout << "is derived2";
  dynamic_cast<derived2_base*>(foo)->specific_derived2_function();
}

NOTE: I consider a list of dynamic_cast<> a code smell, and I urge you to rethink your approach.

Answer 4

You could add a virtual method to do a meta-type check of some kind:

class base
{
public:
    ~base(){}
    virtual void foo() {}
    virtual bool isa(const char* type_to_test){
          return strcmp(type_to_test,"base")==0;}
};

template <typename T>
class derived1 : public base
{
   virtual void foo() {};
   virtual bool isa(const char* type_to_test){
   return strcmp(type_to_test,"derived1")==0;}
};