casting sender() to wanted type returns nullptr

user4592357

in my event handling function i have such code:

void Widget::mousePressEvent(QMouseEvent *event)
{
	const auto button = event->button();
	if (button == Qt::LeftButton)
	{
		emit leftClicked(getName());
	}
        ...
}

QString Widget::getName() const
{
         // return some string
}

then, somewhere else i connect the signal to a slot, where i use the passed string.
now, if i write the above code as

void Widget::mousePressEvent(QMouseEvent *event)
{
	const auto button = event->button();
	if (button == Qt::LeftButton)
	{
		emit leftClicked(); // <---- not passing the argument
	}
        ...
}

and in the connected slot do this

void OtherWidget::onLeftClicked()
{
     if (auto widget = qobject_cast<Widget *>(sender())) // <--- widget == nullptr
         // use widget->getName() analogously 
}

what's wring here?

mrjj

Hi
The type of sender() cannot be altered by omitting a parameter.
Something else must be going on.

Do you connect onLeftClicked to any other object than a Widget? ( bad name btw :)

user4592357

the actual name is different

and absolutely no, this is what i do

auto pWidget = new Widget;
connect(pWidget, &Widget::leftClicked, [this] { onLeftClicked(); });

mrjj

@user4592357
Ah, its a lambda.
please see
https://forum.qt.io/topic/30568/how-to-get-the-sender-in-a-c-lambda-signal-slot-handler/5

Basically you have to capture the sender manually.

user4592357

@mrjj
oh thanks!
i'm on mobile so can't try to run the code.
if i capture the sender, and call the slot from within lambda, and then get sender in that slot, will that be okay? or i need to use the sender object directly in lambda body?

mrjj

@user4592357
You use the captured sender directly in the lambda.