qobject_cast<QVBoxLayout*>(layout()), is the appropriate cast?

Question

Considering the layout was set in a QWidget with the following code:

setLayout(new QVBoxLayout);

And then it needs to be retrieved (to add more stuff to the layout). This was done with the following code:

QHBoxLayout *hLayoutTime(new QHBoxLayout);
qobject_cast<QVBoxLayout*>(layout())->addLayout(hLayoutTime);

qobject_cast is the appropriate kind of cast to use here?

Answer 1

To avoid unneded casting write this like this:

void YourWidget::setupContents()
{
     QVBoxLayout *vLayout = new QVBoxLayout(this); // effectively this does setLayout(new QVBoxLayout);

     QHBoxLayout *hLayoutTime(new QHBoxLayout);
     vLayout->addLayout(hLayoutTime);
     … … …
}

Answer 2

Looking the code in your current example, Why don't you just get the pointer when creating?

auto *vLayout = new QVBoxLayout;
auto *hLayoutTime = new QHBoxLayout;
vLayout->addlaout(hLayoutTime);

Answering your question, probably the most adequate cast is:

dynamic_cast<QHBoxLayout*>(new QVBoxLayout);

dynamic_cast has several checks and benefits over static_cast, so it is better to use it when possible.