i created one QWidget(Parent). in the inside of the parent widget i created another one QWidget(Child). In the run time i need to remove the child widget. how to do that?
i am not using any layout. i am directly putting in the Parent Widget.
Please Help me to fix this.
If you add the widget with e.g.:
QWidget *w = new QWidget(parent);
...then you can remove it with:
delete w;
Another approach would be to just hide it:
w->hide();
This answer is for those arriving from search engines and want an answer to the question as stated in the title.
If you want to remove a child from a parent without deleting it or hiding it (which does NOT remove it from its parent), set the child's parent to NULL.
QWidget::setParent(NULL)
Note that explicitly reparenting a widget like this carries several implications (e.g visibility automatically set to hidden). See QWidgets documentation for more information.
Related
From this post here, in general:
All QObjects will delete their own child objects automatically. (See
docs here.) QWidgets are QObjects. So as long as you establish a
parent/child relationship, you do not need to manually delete your
objects. To do that, simply pass a pointer to the parent object to the
constructor:
QLabel *label1 = new QLabel; // <<- NEED TO DELETE
QLabel *label2 = new QLabel(some_parent_obj); // Will be deleted when some_parent_obj is deleted
So some questions arises:
Does every widget necessary needed/required a parent? If no, what are the exceptions? If yes, what happens to widgets without parent?
I asked this because from examples in Qt Docs, some example widgets have parents (QLabel example) but some doesn't (QBarChart example, and also QFont, QColor, etc...).
So I'm wondering if there's an exception, or those widgets just don't need any parents, or if I declare them with new for some reason, I have to delete afterward.
And vice versa...
Does a widget without parent guarantee to cause a memory leak (or something similar) when the widget which it stays in (not necessary its parent) is deleted? Or if it's removed from a layout without any deletion happening?
Because from my experience with my code, I've created probably quite a lot (~100) of widgets and other stuffs without neither setting any parent (nor using delete afterward), and the project appears to run fine without any stalls even after a while (the effect might be underlying though, as I haven't run Memcheck), hence this question is here.
Does every widget necessary needed/required a parent?
If you want them to be deleted automatically - yes. But...
If no, what are the exceptions? If yes, what happens to widgets without parent?
You do not need to provide a parent to widget if you attach it to layout using QLayout::addWidget. If you look into source code, you'll see that when you do so, it automatically attaches layout's parent as widget's parent(unless you didn't attach layout to any widget).
But if you leave the widget created with new without parent and do not attach to anything - it is leaking memory. You must delete it either using delete or QObject::deleteLater. The last option is recommended when object has any connections.
Does a widget without parent guarantee to cause a memory leak (or something similar) when the widget which it stays in (not necessary its parent) is deleted? Or if it's removed from a layout without any deletion happening?
As I already mention QLayout::addWidget sets parent for added widget, so the answer is no. Also note, that when you call QLayout::removeWidget, it removes only QLayoutItem from layout, but widget's parent stays the same as it was after calling QLayout::addWidget.
(I use Qt 4.7, Windows 7, 64bit).
I created a custom table. Each row is a horizontal layout with widgets.
The rows are kept in a QList for easy access, and the children too. The rows are also added inside the parent widget.
If I resize the parent widget, I calculate the new sizes, delete everything, and recreate it again.
My problem is that I don't want to delete any widget. Only when I clear the table, I do it.
Since I have the widgets inside a QList and inside the parent layouts, How can I remove all widgets in each row, delete all layouts, and then add those to new layouts?
If I do: takeAt(0) for every element inside each layout I have a QLayoutItem with a widget inside... How can I delete the layoutItem without deleting the widget?.... How do I remove the widget without killing it, no matter if it's in the parent or the child? Because there are many methods for deleting: removeItem, removeWidget... in a layout, but not takeWidget... just takeAt() and it gives a Qlayoutitem.
I tried several ways, but I still see the widgets no matter what happened to them.
Questions about this:
When does a widget get deleted? If I takeWidget(index) from a layout, is it deleted some time by itself? does it happen if I have a pointer to it in another list?
removeAt(index) does execute the delete method of a widget?
Ok. I got it working.
Let me explain how this Removing, keeping widgets works.
A widget is known by its parent layout. And you remove it through the layout. By doing:
layout()->removeAt(widget);
delete widget;
If you use takeAt(index) in a QLayout (or its children), it gives you a QLayoutItem. To access the widget inside, just use widget(). But there's no way to remove the widget without deleting it. So this approach is non valid.
In the Docs it tells a way to delete the elements:
QLayoutItem *child;
while ((child = layout->takeAt(0)) != 0) {
...
delete child;
}
A special thing to note in Qt is the following:
If you have a hierarchy tree of layouts, added with addLayout() inside layouts, no matter how deep your widget is inserted, you can remove it from the child layouts or any of the parent layouts, if the tree path from the layout and this item is built from child layouts.
The easiest thing is to keep a list of pointers to all the items, in a custom table. When clearing the table to reconstruct it, just do this inside your widget:
CustomTableItem* item;
while ( !items_.isEmpty() && ( (item = items_.takeFirst()) != 0 ) ){
layout()->removeWidget(item);
delete item; // It works no matter where the item is
}
items_.clear(); // clear the list afterwards.
And it works perfectly, updates the layout too by itself.
If you want to keep the elements, just skip the "delete item;" and use them afterwards.
An important thing to note is that different "remove" functions work differently (as i understand on Qt Docs) in QList or similar widgets, and in a QLayout.
In the QList, removeAt actually removes the object.
(Qt 4.7 QList Docs)"Removes the item at index position i. i must be a valid index position in the list (i.e., 0 <= i < size())."
In a QLayout, removeWidget or removeItem don't remove the item/widget, you have the responsability to delete it, as I did before.
(Qt 4.7 QLayout Docs) "Removes the widget widget from the layout. After this call, it is the
caller's responsibility to give the widget a reasonable geometry or to
put the widget back into a layout."
Hope it helps. If you see any error, you could tell me and I will edit the answer!
More on deleting here:
Other stackoverflow post
A widget in Qt is a regular C++ object and can be deleted with the C++ delete operator as any other object:
delete myWidget;
In Qt there is always a parent-child relation between widgets. When the parent widget is destroyed, it will delete all its children. Usually, you do not need to think about explicitly deleting any widgets but the top level widgets, i.e., windows and dialogs. Qt will take care of deleting any child widgets.
QList::removeAt(int) does not delete the object that is removed, it only removes the object from the list. If you also want to delete the object you would have to do something like:
delete myList.takeAt(0);
This applies to all functions such as removeAt(int), takeAt(int), takeFirst(), etc. They never delete objects, they only remove them from the container (list, layout, scrollarea, etc). In most cases the ownership of the widget is then transferred to the caller, (the caller becomes responsible for deleting the widget as the parent-child relation breaks), but do not assume this is always the case, always read the documentation of the function.
This is driving me nuts. I have a custom menu class that, when set visible, shows a list of items located in a particular folder. When a hardware button is pressed, my application gets the latest list of items, populates the menu with them, and returns.
The menu displaying these items uses a QListWidget filled with custom widgets. Each of the widgets contains one or more QLabels in a horizontal layout, and is created at the time the menu is shown. In order to adjust the text displayed based on the menu width available, I need to get the size of the QLabel AFTER it has been resized according to the layout, but before the menu becomes visible to the user. The problem is, my layout does not get updated until all of the functions constructing my list return.
I have tried QApplication::ProcessEvents() and the layout update functions, but none of them have updated the values of my QLabels before returning. I can set a QTimer when the button is initially pressed, and have it show the menu, update the items, and stop itself, but that seems like a terrible solution.
Any help would really be appreciated! I've spent most of a day on this.
Marlon
I had this exact problem and could not find an answer anywhere on the Internet. Calling Layout.update(), Layout.activate(), or widget.adjustSize() (all suggested in various places) all did not work.
I had a widget with a vertical layout that I wanted to add a QLabel to and then immediately use the size of the QLabel.
The only thing that worked reliably was
layout->addWidget(myLabel);
myLabel->show();
size = myLabel->size();
It would seem that layouts will just not recalculate until you either return from a function and allow the Qt event loop to progress or manually call show() yourself.
How to update a QLayout and get the new dimensions before returning?
Don't. You're not meant to do that. It'll drive you "nuts" because you're doing it backwards. Layout updates are handled asynchronously from the event loop. Instead of getting layout dimensions right away, set yourself up to be part of the system. Some options are:
Implement a custom widget that will interact properly with the layout, growing to fill the available width of the layout. Perhaps all you need is a size policy and a way to elide text?
Make a custom layout that takes the special properties of your use case into account.
You want to call QWidget::adjustSize() on your parent widget. This will force the layout recalculations.
Have you tried using layout()->update(); ?
I've tried many but nothing works for me on Qt 5.15.
Only invented little patch - create timer and get size after 20 msec:
QTimer::singleShot(20, this, [this]
{
const auto height = myLayout->contentsRect().height();
// ...
});
I'm creating an application which has a "floating" widget which can be dragged around inside the application window. But it starts up, or tends to go behind other widgets sometimes. Is there any way to make sure that the widget in my application stays on top of all other widgets whenever it is made visible?
Thanks.
Use the flag Qt::WindowStaysOnTopHint for your QWidget. This will force your widget to stay on top of all other windows
You can call raise() on your widget to make it appear in front of all other child widgets of the parent it is in. If I read your question correctly, this is the behavior you want. However, any child you create and add to a parent widget will automatically be placed above that widget, so you may need to reraise the widget after additions, or you may want to consider an alternate way of managing the parent/child relationship.
In my app have a window splitted by a QSplitter, and I need to remove an widget.
How can I do that? I can't find useful methods
It's not clear to me if you want to preserve the widget and put it somewhere else, or if you want to destroy the widget.
Destroying the widget: If you can
get a pointer to the widget, you can
simply delete it. The splitter will
safely be notified that its child is
being deleted and will remove it
from itself.
Preserving the widget: If you grab
the pointer to the widget, you can
simply set its parent to some other
widget and add it to another
widget's layout and it will show up
there. This is safe because the
QSplitter will be notified that one
of its children is being reparented.
If you want to set the parent to NULL (cjhuitt's answer) be aware that you are now responsible for cleaning up that memory because the widget no longer has a parent.
Many things in Qt cannot be "traditionally" removed. Instead call hide() on it and destruct it. From QSplitter documentation:
When you hide() a child its space will
be distributed among the other
children. It will be reinstated when
you show() it again.
I like Tuminoid's answer. But if you absolutely need it removed, try getting the widget you want to remove, and calling setParent( NULL ) on that widget. That's my best guess.
If you hold a pointer to the widget, then just delete it, or use deleteLater() if you want to be on the safe side.
If you do not have the widget pointer, use QSplitter::widget(int index) function. Then, you can use invoke its deleteLater() slot.
If you do not have the widget index, but you still know the widget objectName(), then QObject::findChild() is your only way to get the widget pointer.
I ran into the same problem. In Qt 4.8 to temporally hide one of the widget of a QSplitter I simply hide it. However it is not enough, as the splitter handle is still available to move. But the handle can be accessed and hidden as well:
frameA->setVisible(conditionA);
frameB->setVisible(conditionB);
if ( !(conditionA && conditionB) ) // if only 1 frame is visible
{
splitter->handle(0)->setVisible(false);
}
Another easy way to prevent the child widget from getting deleted is to use QSplitter.takeWidget(child). This is also the recommended way of removing the widget from a splitter. (Qt Documentation)