Qt 使用 lambda 表达式做为槽函数时为什么使用 QObject::sender() 获取到的发送信号对象指针为空？

/*!

    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 表达式做槽函数时，相当于使用了 Qt::DirectConnection 方式连接槽，都在同一个线程中，故获取发送信号对象指针时，直接给你返回了 0.

巴特西

Qt 使用 lambda 表达式做为槽函数时为什么使用 QObject::sender() 获取到的发送信号对象指针为空？

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