C++ Qt 接口或抽象类和 qobject_cast()
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Qt interfaces or abstract classes and qobject_cast()
提问by Timothy Baldridge
I have a fairly complex set of C++ classes that are re-written from Java. So each class has a single inherited class, and then it also implements one or more abstract classes (or interfaces).
我有一组相当复杂的 C++ 类，它们是从 Java 重写的。所以每个类都有一个继承的类，然后它也实现了一个或多个抽象类（或接口）。
Is it possible to use qobject_cast()to convert from a class to one of the interfaces? If I derive all interfaces from QObject, I get an error due to ambiguous QObjectreferences. If however, I only have the base class inherited from QObject, I can't use qobject_cast()because that operates with QObjects.
是否可以用于qobject_cast()从类转换为接口之一？如果我从 派生所有接口QObject，由于QObject引用不明确，我会收到错误消息。但是，如果我只有从 继承的基类QObject，则无法使用，qobject_cast()因为它与QObjects 一起操作。
I'd like to be able to throw around classes between plugins and DLLs referred to by their interfaces.
我希望能够在插件和它们的接口引用的 DLL 之间抛出类。
回答by Venemo
After some research and reading the qobject_cast documentation, I found this:
经过一番研究和阅读qobject_cast 文档后，我发现了这一点：
qobject_cast() can also be used in conjunction with interfaces; see the Plug & Paint example for details.
qobject_cast() 也可以与接口结合使用；有关详细信息，请参阅即插即用示例。
Here is the link to the example: Plug & Paint.
这是示例的链接：Plug & Paint。
After digging up the interfaces headerin the example, I found the Q_DECLARE_INTERFACEmacro that should let you do what you want.
在挖掘示例中的接口标头后，我找到了Q_DECLARE_INTERFACE宏，它应该可以让你做你想做的事。
First, do notinherit QObjectfrom your interfaces. For every interface you have, use the Q_DECLARE_INTERFACE declaration like this:
首先，不要QObject从你的接口继承。对于您拥有的每个接口，请使用 Q_DECLARE_INTERFACE 声明，如下所示：
class YourInterface
{
public:
virtual void someAbstractMethod() = 0;
};
Q_DECLARE_INTERFACE(YourInterface, "Timothy.YourInterface/1.0")
Then in your class definition, use the Q_INTERFACESmacro, like this:
然后在您的类定义中，使用Q_INTERFACES宏，如下所示：
class YourClass: public QObject, public YourInterface, public OtherInterface
{
Q_OBJECT
Q_INTERFACES(YourInterface OtherInterface)
public:
YourClass();
//...
};
After all this trouble, the following code works:
经过所有这些麻烦，以下代码有效：
YourClass *c = new YourClass();
YourInterface *i = qobject_cast<YourInterface*>(c);
if (i != NULL)
{
// Yes, c inherits YourInterface
}
