Clarification on deleting layouts and widgets within them

[Sh1gs]

Hello Qt Universe,

I would like some clarification and some help on something I'm working on. Let's say that I have the following:
note: this wasn't copied from my code so please forgive syntax errors

QPushButton *btn1, *btn2, *btn3, *btn4, *delBtn;
QHBoxLayout *hbox1, *hbox2;
QVBoxLayout *vbox1;

btn1 = new QPushButton(Tr("1");
btn2 = new QPushButton(Tr("2");
btn3 = new QPushButton(Tr("3");
btn4 = new QPushButton(Tr("4");
delBtn = new QPushButton(Tr("del");

connect(delBtn, &QPushButton::clicked, this, &MyClass::delBtnClicked);

hbox1 = new QHBoxLayout;
hbox1->addWidget(btn1);
hbox1->addWidget(btn2);

hbox2 = new QHBoxLayout;
hbox2->addWidget(btn3);
hbox2->addWidget(btn4);

vbox1 = new QVBoxLayout;
vbox1->addLayout(hbox1);
vbox1->addLayout(hbox2);

setLayout(vbox1);

void MyClass::delBtnClicked()
{
     //remove and delete entire vbox1 layout
}

I've seen the following:

QLayoutItem *child;
while ((child = layout->takeAt(0)) != 0) {
...
delete child;
}

Would each hbox be "child" and "layout" would be vbox? And if I delete "child" will that then delete the QPushButton widgets within hbox1 and hbox2?

Thanks!

[SGaist]

Hi,

Most likely yes.

[Sh1gs]

Well I got my example to work by using the following:

while(!vbox1->isEmpty()) {
    QLayout *hb = vbox1->takeAt(0)->layout();
    while(!hb->isEmpty()) {
        QWidget *w = hb->takeAt(0)->widget();
        delete w;
    }
    delete hb;
}
delete vbox1;

This removes the widgets and the layout they're on, as well as deletes them from memory. Thanks!

[mrjj]

@Sh1gs
Yes Deleting QLayoutItem seems not to free any Widget it manages.

[SGaist]

No, it doesn't. The QLayoutItem is not the parent of the widget, it is not a QObject. That's not its job.