QThread::quit() does not call workerthread's destractor

[ynakane]

Hi.
I'm a beginner of Qt multithreaded programming.

I wrote the multithread code with reference to the Qt reference.

But I noticed something strange.
Workerthread's destructor does not called after QThread :: quit () on main thread.

Here is my code snippet.

I suspect I'm using connect incorrectly.
Thanks for advice.

// Main Thread's source file
MemoryInformation::MemoryInformation(QObject *parent)
    : QAbstractListModel(parent)
{
    // omit
}

MemoryInformation::~MemoryInformation()
{
　// Destractor is called by GUI operation (ex: Close Button)
    m_thread.quit();
    m_thread.wait();
}

void MemoryInformation::ThreadStart()
{
    // This func is called by GUI operation

    Worker* pWorker = new Worker(this);

    // m_thread: QThread Type
    pWorker ->moveToThread(&m_thread);

    connect(&m_thread, &QThread::started, pWorker , &Worker::process);
    connect(pWorker , &Worker::sendElement, this, &MemoryInformation::addElement, Qt::BlockingQueuedConnection);
    connect(&m_thread, &QThread::finished, pWorker , &QObject::deleteLater, Qt::QueuedConnection);

    m_thread.start();
}

void MemoryInformation::addElement(Memory *element)
{
    int pos = elements.count();
    beginInsertRows(QModelIndex(), pos, pos);
    elements.insert(pos, *element);
    endInsertRows();
}

// Sub Thread's source file
Worker::Worker(MemoryInformation *pMemoryInfo, QObject *parent) : QObject(parent)
{
    m_pMemoryInfo = pMemoryInfo;
}

Worker::~Worker()
{
// This function is not called
    delete m_pMemoryInfo;
}

void Worker::process()
{
    check();

    auto timer = new QTimer(this);
    connect(timer, &QTimer::timeout, this, &Worker::check);
    timer->start(1000);
}

void Worker::check()
{
    Memory *element = new <Memory>();

    // omit

    emit sendElement(element);
}

[jsulm]

@ynakane Why should the destructor be called if you terminate a thread? The QThread object still exists, destructor will be called as soon as the QThread instance is destroyed.

[ynakane]

@jsulm Thanks for your comment.
This problem has been resolved
Thank you.