How can I declare a template pointer without knowing the type?

Question

This is what I would like to do:

ExampleTemplate* pointer_to_template;
cin >> number;
switch (number) {
case 1:
    pointer_to_template = new ExampleTempla         


        

Answer 1

You can't. ExampleTemplate<int> and ExampleTemplate<double> are two different, unrelated types. If you always have a switch over several options, use boost::variant instead.

typedef boost::variant<Example<int>, Example<double>> ExampleVariant;
ExampleVariant v;
switch (number) {
    case 1: v = Example<int>(); break;
    case 2: v = Example<double>(); break;
}
// here you need a visitor, see Boost.Variant docs for an example

Another way is to use an ordinary base class with virtual public interface, but I'd prefer variant.

struct BaseExample {
    virtual void do_stuff() = 0;
    virtual ~BaseExample() {}
};

template <typename T>
struct Example : BaseExample { ... };

// ..
BaseExample *obj;

Answer 2

What you're trying to do is not possible. This is becase your ExampleTemplate class doesn't exist by itself, only exists when you relate it with a type.

You could get that behaviour using inheritance:

1. Define a GeneralExampleTemplate (not a template class).
2. Make ExampleTemplate<T> inherit from GeneralExampleTemplate.
3. That way you can create a GeneralExampleTemplate pointer and assign it with a (for example) ExampleTemplate<int>.

Answer 3

You can do something similar by having your template class derive from a regular class:

#include<iostream>
#include<sstream>
using namespace std;

class ExampleBase{

public:
    virtual ~ExampleBase() {}
    virtual string Get() = 0;
};

template<typename T>
class ExampleTemplate : public ExampleBase{

private:
    T data;

public:
    ExampleTemplate(T t) : data(t){}

    string Get(){        
        stringstream s; s << data;
        return s.str();
    }

};

int main(){

    ExampleBase *base;
    int number;
    cout << "> " << flush; cin >> number;

    switch(number){
        case 1:
            base = new ExampleTemplate<int>(42);
            break;
        case 2:
            base = new ExampleTemplate<double>(3.14);
            break;
        default:
            return 1;
    }

    cout << base->Get() << endl;

    delete base;
    return 0;
}