我对QtPlugin存在严重问题。我尝试从接口构建一个插件,名为dictionary:
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
我在blankdictionary项目中使用如此(blankdictionary.h)声明的blankdictionary类中实现了字典。
.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
blankdictionary.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,它会返回以下错误:
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]
任何想法?
提前致谢!
答案 0 :(得分:2)
另一个问题(以及您在评论中提到的问题)是,您已在Q_OBJECT
类中声明了blankdictionary
宏,但未在QObject
中派生它}。