Qt 使用 lambda 表達式做為槽函數時為什么使用 QObject::sender() 獲取到的發送信號對象指針為空？

/*!
    Returns a pointer to the object that sent the signal, if called in
    a slot activated by a signal; otherwise it returns 0. The pointer
    is valid only during the execution of the slot that calls this
    function from this object's thread context.

    The pointer returned by this function becomes invalid if the
    sender is destroyed, or if the slot is disconnected from the
    sender's signal.

    \warning This function violates the object-oriented principle of
    modularity. However, getting access to the sender might be useful
    when many signals are connected to a single slot.

    \warning As mentioned above, the return value of this function is
    not valid when the slot is called via a Qt::DirectConnection from
    a thread different from this object's thread. Do not use this
    function in this type of scenario.

    \sa senderSignalIndex(), QSignalMapper
*/

QObject *QObject::sender() const
{
    Q_D(const QObject);

    QMutexLocker locker(signalSlotLock(this));
    if (!d->currentSender)
        return 0;

    for (QObjectPrivate::Connection *c = d->senders; c; c = c->next) {
        if (c->sender == d->currentSender->sender)
            return d->currentSender->sender;
    }

    return 0;
}

使用 lambda 表達式做槽函數時，相當於使用了 Qt::DirectConnection 方式連接槽，都在同一個線程中，故獲取發送信號對象指針時，直接給你返回了 0.