Specify a base classes template parameters while instantiating a derived class?
- by DaClown
Hi,
I have no idea if the title makes any sense but I can't find the right words to descibe my "problem" in one line. Anyway, here is my problem. There is an interface for a search:
template <typename InputType, typename ResultType> class Search {
public:
virtual void search (InputType) = 0;
virtual void getResult(ResultType&) = 0;
};
and several derived classes like:
template <typename InputType, typename ResultType>
class XMLSearch : public Search<InputType, ResultType> {
public:
void search (InputType) { ... };
void getResult(ResultType&) { ... };
};
The derived classes shall be used in the source code later on. I would like to hold a simple pointer to a Search without specifying the template parameters, then assign a new XMLSearch and thereby define the template parameters of Search and XMLSearch
Search *s = new XMLSearch<int, int>();
I found a way that works syntactically like what I'm trying to do, but it seems a bit odd to really use it:
template <typename T> class Derived;
class Base {
public:
template <typename T>
bool GetValue(T &value) {
Derived<T> *castedThis=dynamic_cast<Derived<T>* >(this);
if(castedThis)
return castedThis->GetValue(value);
return false;
}
virtual void Dummy() {}
};
template <typename T> class Derived : public Base {
public:
Derived<T>() {
mValue=17;
}
bool GetValue(T &value) {
value=mValue;
return true;
}
T mValue;
};
int main(int argc, char* argv[])
{
Base *v=new Derived<int>;
int i=0;
if(!v->GetValue(i))
std::cout<<"Wrong type int."<<std::endl;
float f=0.0;
if(!v->GetValue(f))
std::cout<<"Wrong type float."<<std::endl;
std::cout<<i<<std::endl<<f;
char c;
std::cin>>c;
return 0;
}
Is there a better way to accomplish this?
