在地图中获取不了解模板参数的派生类的模板参数类型

Question

I am trying to get the template parameter type in this scenario:

#include <iostream>
#include <string>
#include <map>
#include <typeinfo>

class Base {
public:
    typedef char myType;
};

template <typename T>
class Derived : public Base {
public:
    typedef T myType;
};

int main() {
    std::map<std::string, Base*> myMap;
    myMap["test1"] = new Derived<int>();
    myMap["test2"] = new Derived<float>();

    std::cout << typeid(myMap["test1"]).name() << std::endl; // prints Base
    std::cout << typeid(myMap["test2"]).name() << std::endl; // prints Base

    //myMap["test1"]->myType test; // invalid use of 'Base::myType'

    std::cout << typeid(dynamic_cast<Derived*>(myMap["test1"])->myType).name() << std::endl; // invalid use of template-name 'Derived' without an argument list. Should print "int" ...
    std::cout << typeid(dynamic_cast<Derived*>(myMap["test2"])->myType).name() << std::endl; // invalid use of template-name 'Derived' without an argument list. Should print "float" ...
}

The map holds elements of type Base and thus, also of type Derived with a template parameter. However on retrieving elements from the map I am not able to get the template parameter type back again. I tried to add a typedef to both classes but it doesn't work.

Do you have hints to resolve this problem?

Thanks in advance!

Answer 1

Do you have hints to resolve this problem?

type names don't work like virtual member functions. What you need is a virtual member function.

Here's a demonstrative program:

#include <iostream>
#include <string>
#include <map>
#include <typeinfo>

class Base {
public:
    virtual std::type_info const& myType() const { return typeid(char); }
};

template <typename T>
class Derived : public Base {
public:
    virtual std::type_info const& myType() const { return typeid(T); }
};

int main() {
    std::map<std::string, Base*> myMap;
    myMap["test1"] = new Derived<int>();
    myMap["test2"] = new Derived<float>();

    std::cout << myMap["test1"]->myType().name() << std::endl;
    std::cout << myMap["test2"]->myType().name() << std::endl;
}

Output with g++:

i
f