我有一个严重的QtPlugin问题。我试图从一个接口,一个名为词典编写一个插件:QT插件编译问题
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blankdictionary项目实施的字典中声明如下(blankdictionary.h)一blankdictionary类。
的.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
的blankdictionary.h文件:
class blankdictionary : public dictionary {
Q_OBJECT
Q_INTERFACES(dictionary)
public:
blankdictionary();
void collectData();
void collectOperator();
void collectControl();
};
在我blankdictionary.cpp文件的末尾,我有:
QT_BEGIN_NAMESPACE
Q_EXPORT_PLUGIN2(blanktarget, blankdictionary)
QT_END_NAMESPACE
在编译,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]
有什么想法吗?
在此先感谢!
我认为问题来自这里: “[...]使通过插件的应用程序扩展包括以下步骤: 定义一组接口(只有纯虚函数的类)用于交谈的插件[...]“ 来自此源:http://doc.qt.nokia.com/4.7-snapshot/plugins-howto.html。 所以,我认为我必须让字典界面只有纯虚函数的类。 我现在试试。 – ocds