c++ - 将 QObject 接口(interface)信号连接到 lambda 插槽

Question

我正在尝试将 QObject 信号连接到 lambda 槽，但使用指向对象的接口(interface)指针而不是指向具体 QObject 类的指针。但是我得到了这个奇怪的错误:

 error: no matching function for call to ‘FileSystemModel::connect(model_filesystem::Directory*&, const char*, FileSystemModel::setDirectory(model_filesystem::Directory*)::<lambda()>)’
 });

这是我的一些代码片段:

// Interface declaration
namespace model_filesystem {

class Directory {
public:

    virtual ~Directory()
    virtual QString name() = 0;

Q_SIGNALS:
        void changed();
        void failure(QString msg);
    };
}

Q_DECLARE_INTERFACE(model_filesystem::Directory, "org.moonlightde.panel.model_filesystem.Directory/1.0")

//Implementation
class GVFSDirectory : public QObject,  public model_filesystem::Directory {
    Q_OBJECT
    Q_INTERFACES(model_filesystem::Directory)
public:
    GVFSDirectory(const QString &uri);
    GVFSDirectory(GFile * gfile);

    virtual ~GVFSDirectory();

    virtual QString name();
public Q_SLOTS:
    void update();

Q_SIGNALS:
    void changed();
    void failure(QString msg);

// Usage
Directory * directory = new GVFSDirectory("/");
connect(directory, SIGNAL(model_filesystem::Directory::changed()), [this] () {
    setupModel();
});

Answer 1

error: no matching function for call to ‘FileSystemModel::connect(model_filesystem::Directory ...

Directory 类不是QObject。如果要使用信号和/或槽，它需要子类QObject。

来自 Qt5 - Signals & Slots :

All classes that inherit from QObject or one of its subclasses (e.g., QWidget) can contain signals and slots.

来自 Qt5 - Plug & Paint Example :

To make it possible to query at run-time whether a plugin implements a given interface, we must use the Q_DECLARE_INTERFACE() macro.

这似乎提供了在运行时查询插件是否实现了给定接口(interface)的机制，所以我不明白为什么应该对 Directory 有任何期望类应该像 QObject 一样工作，而无需对其进行子类化。

换句话说，不用继承 QObject 就可以使用它，但这不会授予 Directory 使用信号和/或插槽。

SIGNAL(model_filesystem::Directory::update())

即使 Directory 是一个 QObject 也没有 update 信号，只有 changed 和 failure 。