Finding template argument of child

[gao_bolin]

I have the following situation:

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

template < typename T >
class B : public A
{
};

The problem is to find the template argument T when a function receives
only a pointer A* (and we know for sure that the A* indeed points to a
B<T>*).

One way is to dynamic_cast to a bunch of know B<T>, but this is kind of
heavy and works only for types that are explicitely taken into account.
Is there a way to retrieve type T for any T, and say print it out?

Thanks

Bolin

 

[Ruslan Abdikeev]

template < typename T >
class B : public A {};

The problem is to find the template argument T when a function receives
only a pointer A* (and we know for sure that the A* indeed points to a
B<T>*).

In which context do you want to find the T?
If it is just to "say, print", then the easiest way is to define a
non-template intermediate base class which implements the necessary
interface:

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

struct A
{
virtual ~A() {}
};

struct B_base : A
{
std::string get_T_name() const { return vget_T_name(); }
private:
virtual std::string vget_T_name() const =0;
};

template<typename T>
struct B : B_base
{
private:
std::string vget_T_name() const { return typeid(T).name(); }
};

int main()
{
B<int> b_int;
A& a= b_int;
B_base& b_base= dynamic_cast<b_base&>( a );
std::cout << b_base.get_T_name() << std::endl;
}


Hope it helps,
Ruslan Abdikeev.

 

[adbarnet]

How about:
class A

{

public:

A(){}

virtual ~A() {}

virtual const type_info & getTypeInfo() const {return typeid(A); }

};

template <class T>

class B : public A

{

public:


virtual const type_info & getTypeInfo() const {return typeid(T); }

};

int main(int argc, char **argv)

{

A anA;

A* pA = new B<int>;

std::cout << anA.getTypeInfo().name() << std::endl;

std::cout << pA->getTypeInfo().name() << std::endl;

return 0;

}

regards,
Aiden