QT_INTERFACES and signal definitions

Question

How to define signals in interfaces?

class ISomeInterface
{
    signals:
        virtual void added(const QString& id) = 0;
        virtual void removed(const QString& id) = 0;

    // bla, other methods
}

Afaik, signals should not virtual but if I want to define that a signal needs to be implemented in classes implementing the interface => how to do this? Or do you simply make them non-abstract... but interfaces have no Q_OBJECT declaration here! Is the correct code generated in this cases? On top... you'll need a bad (f***g) cast to QObject if you want to connect to the signals.

class ISomeInterface
{
    signals:
        void added(const QString& id);
        void removed(const QString& id);

    // bla, other methods
}

Or do you try to implement it that way?

class ISomeInterface : public QObject
{
    Q_OBJECT

    signals:
        void added(const QString& id);
        void removed(const QString& id);

    // bla, other methods
}

Q_DECLARE_INTERFACE(ISomeInterface, "ISomeInterface")

.. but this way I can inherit from only one interface (QObject does not support multiple inheritance).

Conclusion: As suggested, I would go with the first one.

score 3 · Accepted Answer · edited May 23 '17 at 12:25

3

Define signals as virtual abstract protected methods. (Instead of protected you may use signal: keyword).
Final object must interhit your interface and QObject
Cast pointer to interface to QObject * with dynamic_cast
Connect pointer to necessary slots

You may check this topic for some solutions.

edited May 23 '17 at 12:25

Community

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1

answered Sep 11 '13 at 13:45

Dmitry Sazonov

8,801
1
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score 2 · Answer 2 · edited May 23 '17 at 12:25

2

Unfortunately Qt does not handle multiple inheritance from QObject (see this Q&A), so you can't derive all your interfaces from it (otherwise you won't be able to derive from multiple interfaces). So instead you might have something like IQObject interface with

virtual QObject* qObject() = 0

method, so you don't need to cast them to QObject all the time.

As for providing interface for signals - no such thing. What you can do is make them virtual abstract, that'll work if you really need to enforce this compile-time check (because signals: is just an alias for protected:). If you don't, just check the return value from the connect() method.

edited May 23 '17 at 12:25

Community

1
1

answered Sep 11 '13 at 13:38

spiritwolfform

2,263
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1

The `qObject` method is guaranteed to outperform `dynamic_cast(pInterface)` and thus is a good idea. There are no guarantees about `dynamic_cast`'s complexity, while `qObject` has a constant cost. The implementation `QObject * MyClass::qObject() { return this; }` amounts to simple pointer arithmetic. – Kuba hasn't forgotten Monica Mar 29 '17 at 20:41

QT_INTERFACES and signal definitions

2 Answers2