继承接口实现

Question

我是 C++ 的新手，但有一些高级语言（Java 等）的编程经验。我目前面临的主要问题是理解 C++.

中的继承和接口

testsurface.h

#include <QSurface>

class TestSurface : public virtual QSurface
{
public:
    TestSurface(SurfaceClass clazz) : QSurface(clazz) { }
    ~TestSurface() { }

    virtual QSurfaceFormat      format()        = 0;
    virtual QSize               size()          = 0;
    virtual QPlatformSurface*   surfaceHandle() = 0;
    virtual SurfaceType         surfaceType()   = 0;
};

testwindow.h

#include <QWindow>
#include "testsurface.h"

class TestWindow : public QWindow, public TestSurface
{
public:
    TestWindow() : QWindow(), TestSurface(QWindow::Window) { }
    ~TestWindow() { }
};

据我了解，TestSurface 传递了 QSurface 的所有抽象方法。因为 TestWindow 正在继承 QWindow 而 QWindow 正在继承 QSurface 我希望 TestSurface::QSurface 所有需要的方法都是通过 TestWindow::QWindow 实现的，但事实并非如此因为我在构建时收到以下错误消息：

C2512: C:\...\Qt\Projects\Test2\testwindow.h:10: Erroe: C2512: 'QSurface::QSurface': no appropriate default constructor available

和

C:\...\Qt\Projects\Test2\main.cpp:12: Error: C2259: 'TestWindow': Instance of abstract class can't be created because of following members:
"QSurfaceFormat TestSurface::format(void)": is abstract
    c:\...\qt\projects\test2\testsurface.h(12): See declaration of 'TestSurface::format'
"QSize TestSurface::size(void)": is abstract
    c:\...\qt\projects\test2\testsurface.h(13): See declaration of 'TestSurface::size'
"QPlatformSurface *TestSurface::surfaceHandle(void)": is abstract
    c:\...\qt\projects\test2\testsurface.h(14): See declaration of 'TestSurface::surfaceHandle'
"QSurface::SurfaceType TestSurface::surfaceType(void)": is abstract
    c:\...\qt\projects\test2\testsurface.h(15): See declaration of 'TestSurface::surfaceType'
"QSurfaceFormat QSurface::format(void) const": is abstract
    D:\...\msvc2013_64_opengl\include\QtGui/qsurface.h(67): See declaration of 'QSurface::format'
"QPlatformSurface *QSurface::surfaceHandle(void) const": is abstract
    D:\...\msvc2013_64_opengl\include\QtGui/qsurface.h(68): See declaration of 'QSurface::surfaceHandle'
"QSurface::SurfaceType QSurface::surfaceType(void) const": is abstract
    D:\...\msvc2013_64_opengl\include\QtGui/qsurface.h(70): See declaration of 'QSurface::surfaceType'
"QSize QSurface::size(void) const": is abstract
     D:\...\msvc2013_64_opengl\include\QtGui/qsurface.h(73): See declaration of 'QSurface::size'

Answer 1

那不行，QWindow继承QSurface时没有使用虚拟继承。

class Q_GUI_EXPORT QWindow : public QObject, public QSurface {...};

如果你想要"join a base interface"的多个接口，它们都必须使用虚拟继承。

           Base
            / \
           /   \
  virtual /     \ virtual
     Der1         Der2
          \     /
           \   /
            \ /
           Join

继承接口实现

Inherit Interface Implementation

c++

inheritance

qt