如何将信号转发到私有信号？答案

【问题标题】：How do I forward a signal to private signal?如何将信号转发到私有信号？
【发布时间】：2020-09-28 20:05:28
【问题描述】：

假设我有以下课程：

class Example : public QObject {
    Q_OBJECT
public:
    explicit Example(QObject * parent = nullptr);
    void example();

signals:
    void publicSignal();
    void privateSignal(QPrivateSignal);
};

privateSignals 使用QPrivateSignal，因此只能由各自的类发出。由于 moc 编译器从签名中删除了QPrivateSignal，我可以照常使用connect：

Example * foo = new Example(this);
Example * bar = new Example(this);

connect(foo, &Example::publicSignal,  bar, &Example::example); // works fine, signatures fit
connect(foo, &Example::privateSignal, bar, &Example::example); // works fine, QPrivateSignal not part of signal

我可以轻松地将两个信号变体转发到publicSignal 变体：

connect(foo, &Example::publicSignal,  bar, &Example::publicSignal); // forward foo's signals to bar
connect(foo, &Example::privateSignal, bar, &Example::publicSignal); // forward foo's private signals to bar

这对于私有信号显然是不可能的，即使在 Example's 构造函数中：

Example::Example(QObject * parent) : QObject(this) {
    // Suppose that SimilarExampleClass looks the same but does not recurse
    SimilarExampleClass * other = new SimilarExampleClass(this);

    // this does not compile either
    connect(other, &SimilarExampleClass::publicSignal, this, &Example::privateSignal);
}

这两种情况都会导致以下错误：

$QTHOME/include/QtCore/qobject.h: In instantiation of ‘static QMetaObject::Connection QObject::connect(const typename QtPrivate::FunctionPointer<Func>::Object*, Func1, const typename QtPrivate::FunctionPointer<Func2>::Object*, Func2, Qt::ConnectionType) [with Func1 = void (SimilarExampleClass::*)(); Func2 = void (Example::*)(Example::QPrivateSignal); typename QtPrivate::FunctionPointer<Func>::Object = SimilarExampleClass; typename QtPrivate::FunctionPointer<Func2>::Object = Example]’:
example.cpp:107:89:   required from here
$QTHOME/include/QtCore/qobject.h:239:9: error: static assertion failed: The slot requires more arguments than the signal provides.
         Q_STATIC_ASSERT_X(int(SignalType::ArgumentCount) >= int(SlotType::ArgumentCount),
         ^
$QTHOME/include/QtCore/qobject.h:241:9: error: static assertion failed: Signal and slot arguments are not compatible.
         Q_STATIC_ASSERT_X((QtPrivate::CheckCompatibleArguments<typename SignalType::Arguments, typename SlotType::Arguments>::value),
         ^
...

$QTHOME/include/QtCore/qobject.h: In instantiation of ‘static QMetaObject::Connection QObject::connect(const typename QtPrivate::FunctionPointer<Func>::Object*, Func1, const typename QtPrivate::FunctionPointer<Func2>::Object*, Func2, Qt::ConnectionType) [with Func1 = void (SimilarExampleClass::*)(SimilarExampleClass::QPrivateSignal); Func2 = void (Example::*)(Example::QPrivateSignal); typename QtPrivate::FunctionPointer<Func>::Object = SimilarExampleClass; typename QtPrivate::FunctionPointer<Func2>::Object = Example]’:
example.cpp:108:90:   required from here
$QTHOME/include/QtCore/qobject.h:241:9: error: static assertion failed: Signal and slot arguments are not compatible.
         Q_STATIC_ASSERT_X((QtPrivate::CheckCompatibleArguments<typename SignalType::Arguments, typename SlotType::Arguments>::value),
         ^

有什么方法可以在信号类中使用私有信号作为connect 的目的地？

【问题讨论】：

标签： c++ qt qt5 signals-slots

【解决方案1】：

不幸的是，没有一个小帮手。 Qt 的自动QPrivateSignal 过滤器从其connect 语法中的信号中过滤不移植到它的槽参数。因此，您需要为您的Example 提供一个 lambda：

connect(other, &SimilarExampleClass::publicSignal, this, [this](){
    emit privateSignal({});
});

【讨论】：

【解决方案2】：

老式信号/槽形式确实适用于私有信号：

connect(other, SIGNAL(publicSignal()), this, SIGNAL(privateSignal()));

我在尝试连接到 QTimer 的信号超时时偶然发现了这个问题，至少从 Qt5.9 开始这是一个私有信号

【讨论】：