我正在创建一个Qt接口来强制任何子类实现两个主要的方法集和标题。但是,当我尝试编译它时,我收到一个奇怪的错误消息,它提到了有关qt_check_for_QOBJECT_macro和staticMetaObject的一些信息。 在mainwindow.cpp中,我必须将任何页面转换为接口,以便我可以依赖getter和setter方法。 我没有看到任何其他方式来做到这一点。Qt纯虚函数错误
这是我的代码:
//IPage.h
#ifndef IPAGE_H
#define IPAGE_H
#include <QString>
class IPage
{
public:
virtual QString title()=0;
virtual void setTitle(QString t)=0;
};
#endif // IPAGE_H
//buildings.h:
#ifndef BUILDINGS_H
#define BUILDINGS_H
#include "IPage.h"
#include <QDialog>
class Buildings : public IPage, public QDialog
{
Q_OBJECT
private:
QString m_title;
//stuff...
};
#endif
//buildings.cpp
//stuff...
void Buildings::setTitle(QString t)
{
m_title = t;
setWindowTitle(t);
}
QString Buildings::title()
{
return m_title;
}
//mainwindow.cpp:
QMdiSubWindow *MainWindow::findChild(const QString &title)
{
foreach (QMdiSubWindow *window, mdiArea->subWindowList()) {
IPage *child = qobject_cast<IPage *>(window->widget()); /*line 178*/
if (child->title() == title)
return window;
}
return 0;
}
,当我编译我的代码,我收到此错误信息:
In file included from c:\QtSDK\Desktop\Qt\4.8.1\mingw\include/QtCore/qabstractanimation.h:45,
from c:\QtSDK\Desktop\Qt\4.8.1\mingw\include/QtCore/QtCore:3,
from c:\QtSDK\Desktop\Qt\4.8.1\mingw\include\QtGui/QtGui:3,
from mainwindow.cpp:1:
c:\QtSDK\Desktop\Qt\4.8.1\mingw\include/QtCore/qobject.h: In function 'T qobject_cast(QObject*) [with T = IPage*]':
mainwindow.cpp:178: instantiated from here
c:\QtSDK\Desktop\Qt\4.8.1\mingw\include/QtCore/qobject.h:378: error: 'class IPage' has no member named 'qt_check_for_QOBJECT_macro'
c:\QtSDK\Desktop\Qt\4.8.1\mingw\include/QtCore/qobject.h:380: error: 'class IPage' has no member named 'staticMetaObject'
mingw32-make.exe[1]: Leaving directory `D:/Dropbox/Programmi/Qt/Scadenziario'
mingw32-make.exe[1]: *** [build/o/mainwindow.o] Error 1
mingw32-make.exe: *** [debug] Error 2
01:23:26: The process "C:\QtSDK\mingw\bin\mingw32-make.exe" exited with code 2.
Error while building project scadenziario (target: Desktop)
When executing build step 'Make'
我无法理解的错误消息。我试图谷歌它,但我找不到任何有用的信息。 任何帮助将不胜感激,在此先感谢。
我怎样才能解决这个错误? IPage它只是一个界面。我不需要它从QObject继承。我需要确定任何IPage的实现都有getter和setter方法。 – Al79 2012-07-15 00:32:25
对于使用'qobject_cast','IPage' **必须从'QObject'继承:'class IPage:public QObject'。如果你不想这样做,你将不得不在你的'IPage'类中实现一个新的方法'虚拟IPage * IPage :: fromQObject(QObject * obj)= 0;'(或类似的东西)在C++中不存在)。 –
2012-07-15 00:52:11