c++ - 在接口(interface)类中声明抽象信号

Question

当实现类已经从 QObject/QWidget 派生时，如何在抽象类/接口(interface)中声明 Qt 信号？

class IEmitSomething
{
   public:
     // this should be the signal known to others
     virtual void someThingHappened() = 0;
}

class ImplementEmitterOfSomething : public QWidget, public IEmitSomething
{
     // signal implementation should be generated here
     signals: void someThingHappended();
}

Answer 1

正如我在最后几天发现的...... Qt 这样做的方式是这样的:

class IEmitSomething
{
   public:
     virtual ~IEmitSomething(){} // do not forget this

   signals: // <- ignored by moc and only serves as documentation aid
            // The code will work exactly the same if signals: is absent.
     virtual void someThingHappened() = 0;
}

Q_DECLARE_INTERFACE(IEmitSomething, "IEmitSomething") // define this out of namespace scope

class ImplementEmitterOfSomething : public QWidget, public IEmitSomething
{
   Q_OBJECT
   Q_INTERFACES(IEmitSomething)

   signals:
      void someThingHappended();
}

现在您可以连接到这些接口(interface)信号了。

如果您在连接信号时无法访问实现，您的连接语句将需要动态转换为 QObject:

IEmitSomething* es = ... // your implementation class

connect(dynamic_cast<QObject*>(es), SIGNAL(someThingHappended()), ...);

... 这样您就不必将实现类公开给订阅者和客户。是啊!!!