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Home/ Questions/Q 6832595
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T22:52:39+00:00 2026-05-26T22:52:39+00:00

This is what I would like to do: ExampleTemplate* pointer_to_template; cin >> number; switch

  • 0

This is what I would like to do:

ExampleTemplate* pointer_to_template;
cin >> number;
switch (number) {
case 1:
    pointer_to_template = new ExampleTemplate<int>();
    break;
case 2:
    pointer_to_template = new ExampleTemplate<double>();
    break;
}
pointer_to_template->doStuff();

This doesn’t compile because the template type must be specified when declaring the pointer. (ExampleTemplate* pointer_to_template should be ExampleTemplate<int>* pointer_to_template.) Unfortunately, I don’t know the type of the template until it’s declared in the switch block. What is the best work around for this situation?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T22:52:40+00:00Added an answer on May 26, 2026 at 10:52 pm

    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;
    
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