1

我对 QtPlugin 有一个严重的问题。我尝试从一个名为字典的接口构建一个插件：

class dictionary
{
private:
   ... some private members
public:
   ~dictionary();

   ... some no virtual methods

   virtual void collectData()=0;
   virtual void collectOperator()=0;
   virtual void collectControl()=0;
};

QT_BEGIN_NAMESPACE
Q_DECLARE_INTERFACE(CDictionnary, "shinoe.cameleon.dictionary/2.0")
QT_END_NAMESPACE

我已经在一个空白字典项目中这样声明的空白字典类（blankdictionary.h）中实现了字典。

.pro 文件：

!include(../../../configuration.pri)

QT       += core gui

TARGET = blanktarget
TEMPLATE = lib
CONFIG += plugin

#dictionary interface includes
!include(../../../machine/kernel/includekernel.pri)

SOURCES += blankdictionary.cpp
HEADERS += blankdictionary.h

空白字典.h 文件：

class blankdictionary : public dictionary {
   Q_OBJECT
   Q_INTERFACES(dictionary)
public:
   blankdictionary();
   void collectData();
   void collectOperator();
   void collectControl();
};

在我的 blankdictionary.cpp 文件的末尾，我有：

QT_BEGIN_NAMESPACE
Q_EXPORT_PLUGIN2(blanktarget, blankdictionary)
QT_END_NAMESPACE

在编译时，它返回此错误：

blankdictionary.cpp: In function 'QObject* qt_plugin_instance()':
blankdictionary.cpp:20: error: no match for 'operator=' in '_instance = (operator new(44u), (<statement>, ((blankdictionary*)<anonymous>)))'
c:\QtSDK\Desktop\Qt\4.7.4\mingw\include/QtCore/qpointer.h:65: note: candidates are: QPointer<T>& QPointer<T>::operator=(const QPointer<T>&) [with T = QObject]
c:\QtSDK\Desktop\Qt\4.7.4\mingw\include/QtCore/qpointer.h:67: note:                 QPointer<T>& QPointer<T>::operator=(T*) [with T = QObject]

任何的想法？

提前致谢 ！

4

1 回答 1

2

另一个问题（以及您在评论中提到的问题）是您已经Q_OBJECT在类中声明了宏blankdictionary，但没有从QObject.

于 2012-05-08T12:19:11.803 回答