如何将信号从 qconcurrent 线程连接到共享一个字符串的 gui 线程

Question

我正在尝试使用其他线程信息 (QString) 更新 gui 标签。

class MainWindow : public QMainWindow
{
Q_OBJECT

public:
    explicit MainWindow(QWidget *parent = 0);
    ~MainWindow();

public Q_SLOTS:
    void sl_appendInfo(QString p_text);

private:
    Ui::MainWindow *ui;
    QFuture<void> m_thread;
    QFuture<void> m_engine;
    engine* m_object;
};

MainWindow::MainWindow(QWidget *parent) : QMainWindow(parent), ui(new Ui::MainWindow)
{
    m_object = new engine();

    qRegisterMetaType<QString>();
    bool success = connect(this->m_object, SIGNAL(engine::sig_appendInfo(QString)), this, SLOT(sl_appendInfo(QString)), Qt::QueuedConnection);

    if(!success)
    {
        qDebug("success failed");
    }

    m_engine = QtConcurrent::run(this->m_object, &engine::eventLoop);
}

//slot declaration in mainwindow.cpp
void MainWindow::sl_appendInfo(QString p_text)
{
    ui->label->setText(p_text.toLocal8Bit().constData());
}


class engine : public QObject
{

    Q_OBJECT

public:
    engine();
    ~engine();
    void eventLoop();

Q_SIGNALS:
    void sig_exitengine(void);
    void sig_appendInfo(QString p_text);
};

void engine::eventLoop()
{
    int state = false;

    while(true)
    {
        state = getNextEvent(m_event);

        if (state == true)
        {
                sig_appendInfo("information for gui: we handled a new event !");
        state=false;
        }
        QThread::msleep(1000);
    }
}

现在我使用这个 link : My signal / slot connection does not work 来构建我自己的代码，但是它没有用，连接失败...请问有什么帮助吗？

谢谢

Answer 1

您的 connect 语法错误。您不应在 SIGNAL 宏中包含 class 名称。如果你使用旧的语法，它应该是：

bool success = connect(m_object, SIGNAL(sig_appendInfo(QString)), this, SLOT(sl_appendInfo(QString)), Qt::QueuedConnection);

或者，如果您想使用新语法：

bool success = connect(m_object, &engine::sig_appendInfo, this, &MainWindow::sl_appendInfo, Qt::QueuedConnection);

如何将信号从 qconcurrent 线程连接到共享一个字符串的 gui 线程

How to connect signal from qconcurrent thread to gui thread sharing one string

multithreading

qt

signals

connect

slots