c++ - 在接口(interface)类中声明抽象信号

当已经从QObject/QWidget派生实现类时，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();
}

最佳答案

正如我在最近几天发现的那样... Qt的实现方式是这样的:

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。

如果在连接信号时无法访问实现，则connect语句将需要动态转换为QObject:

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

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

...并且通过这种方式，您不必强制将实现类公开给订阅者和客户端。是的!