QThread 阻塞主应用程序

发布于 2024-09-09 00:16:22 字数 510 浏览 4 评论 0原文

我有一个简单的表单 UI，其中有一个用于启动线程的按钮槽：

void MainWindow::LoadImage()
{
    aThread->run();
}

run() 方法如下所示：

void CameraThread::run()
{
    qDebug("Staring Thread");
    while(1)
    {
        qDebug("ping");
        QThread::sleep(1);
    }
}

当我单击调用 LoadImage() 的按钮时，UI 变得无响应。我定期看到“ping”消息作为调试输出，但 UI 挂起，不响应任何内容。为什么我的线程没有单独运行？ CameraThread 派生为公共 QThread 我正在使用 gcc 版本 4.4.3 (Ubuntu 4.4.3-4ubuntu5) 以及 Ubuntu 10.04(x86) 存储库中的 QT 库和 QT Creator。

原文

I have a simple form UI that has a slot for a button, starting a thread:

void MainWindow::LoadImage()
{
    aThread->run();
}

And the run() method looks like this:

void CameraThread::run()
{
    qDebug("Staring Thread");
    while(1)
    {
        qDebug("ping");
        QThread::sleep(1);
    }
}

When I click the button that calls LoadImage(), the UI becomes unresponsive. I periodically see the "ping" message as the debug output but the UI hangs, does not respond to anything.
Why is my thread not running separately? CameraThread derived as public QThread
I am using gcc version 4.4.3 (Ubuntu 4.4.3-4ubuntu5) with QT libraries and QT Creator from Ubuntu 10.04(x86) repositories.

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我偏爱纯白色 2024-09-16 00:16:22

简短回答：通过调用 aThread->start(); 而不是 run() 来启动线程，并确保线程是 run()方法受保护（非公开）。

说明

调用start()是启动线程的正确方法，因为它提供优先级调度并实际上在其自己的线程上下文中执行 run() 方法。

看起来您将在此线程中加载图像，因此我将在您遇到许多人在使用 QThread 时陷入的陷阱之前提供一些提示

QThread 本身不是线程。只是一个线程的包装，这给我们带来了..
CameraThread 类中定义的信号/槽不一定会在线程的上下文中运行，只记住 run( ) 方法和从它调用的方法在单独的线程中运行。

恕我直言，在大多数情况下子类化 QThread 并不是正确的选择。您可以使用以下代码更简单地完成此操作，并且它将为您省去很多麻烦。

class ImageLoader : public QObject {
Q_OBJECT
public slots:
    void doWork() 
    {
        // do work
    }
};

void MainWindow::MainWindow(/*params*/) 
{
  ImageLoader loader;
  QThread thread;
  loader.moveToThread( &thread );
  connect( this, SIGNAL( loadImage() ), &loader ,SLOT( doWork() ) );
  thread.start();
  // other initialization
}
void MainWindow::LoadImage()
{
   emit loadImage();
}

另请阅读有关此主题的 Qt 博客。

Short answer: Start your thread by calling aThread->start(); not run(), and make sure you thread's run() method is protected (not public).

Explanation

Calling start() is the correct way to start the thread, as it provides priority scheduling and actually executes the run() method in its own thread context.

It looks like you are going to be loading images in this thread, so I'm going to include some tips before you run into pitfalls many people fall into while using QThread

QThread itself is not a thread. It is just a wrapper around a thread, this brings us to..
signals/slots defined in the CameraThread class will not necessarily run in the thread's context, remember only the run() method and methods called from it are running in a separate thread.

IMHO, subclassing QThread in the majority of cases is not the way to go. You can do it much simpler with the following code, and it will save you many headaches.

class ImageLoader : public QObject {
Q_OBJECT
public slots:
    void doWork() 
    {
        // do work
    }
};

void MainWindow::MainWindow(/*params*/) 
{
  ImageLoader loader;
  QThread thread;
  loader.moveToThread( &thread );
  connect( this, SIGNAL( loadImage() ), &loader ,SLOT( doWork() ) );
  thread.start();
  // other initialization
}
void MainWindow::LoadImage()
{
   emit loadImage();
}

Also read the Qt blog regarding this topic.

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