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I would like to use a template class to provide some common functionality to some child classes that are very similar. The only variation is the enumeration that each uses.

This is the parent class

template<typename T> class E_EnumerationBase : public SimpleElement
{
public:
    E_EnumerationBase();
    virtual bool setValue(QString choice);
    virtual T getState();

protected:
    T state;
    QHash<QString, T> dictionary;
};

template<typename T> E_EnumerationBase<T>::E_EnumerationBase() {
    state = 0;
}

template<typename T> bool E_EnumerationBase<T>::setValue(QString choice) {
    T temp;
    temp = dictionary.value(choice, 0);
    if (temp == 0) {
        return false;
    }

    value = choice;
    state = temp;
    return true;
}

template<typename T> T E_EnumerationBase<T>::getState() {
    return state;
}

This is one of the children

enum TableEventEnum {
    NO_VALUE = 0,
    ATTRACT = 1,
    OPEN = 2,
    CLOSED = 3
};

class E_TableEvent : public E_EnumerationBase<enum TableEventEnum>
{
public:
    E_TableEvent();
};

This is the constructor

E_TableEvent::E_TableEvent()
{
    state = NO_VALUE;
    dictionary.insert("attract", ATTRACT);
    dictionary.insert("open", OPEN);
    dictionary.insert("closed", CLOSED);
}

The linker is throwing this error:

e_tableevent.cpp:6: error: undefined reference to `E_EnumerationBase<TableEventEnum>::E_EnumerationBase()'

Can an enumeration be used as the parameter to a template like this?

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1 Answer

Enumerations can be template parameters in exactly the same way that ints can.

enum Enum { ALPHA, BETA };

template <Enum E> class Foo {
    // ...
};

template <> void Foo <ALPHA> :: foo () {
    // specialise
}

class Bar : public Foo <BETA> {
    // OK
}

But you simply haven't provided a definition for E_EnumerationBase::E_EnumerationBase()

This isn't a problem with templates or inheritence. It's the same as if you written this:

struct Foo {
    Foo ();
}
int main () {
    Foo foo;
}

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