Removing widget from layout

[lotek]

Hello,

I'd like to remove a widget from the layout and delete it.

for this I'd like to use the following source code

...
int cnt;
QLayoutItem *it;
MyWidget *w;

cnt = aVLayout->count();

for( int i = 0; i < cnt; i++ )
{
it = aVLayout->itemAt(0);
w = ((MyWidget *)(it->widget()));
aVLayout->removeWidget( w );
delete w;
aVLayout->removeItem( it );
delete it; // SEGMENTATION VIOLATION CRASH!!!
}

My questions:
- Is it necessary to remove and delete the QLayoutItem item?
- how to do it to avoid the crash?

In my understanding, the QLayoutItem has been created and added, when the addWidget() method has been called, so when I am removing a widget from the layout and deleting it, also the QLayoutItem shall be removed and deleted. Is it correct?

I am using Qt 4.6.3

Thx for your help

[DanH]

I suspect you should remove the item first, then delete it. Since it owns the widget, there would be no need to separately delete the widget. In your code you've deleted the widget while the item still "owns" it, so the item destructor attempts to delete the widget again.

[SixDegrees]

This approach can never work; you're guaranteed to walk off the end of the layout vector once you remove something from it, because you keep looping over the entire initial count before removal.

[Zlatomir]

A simple approach for this kind of "dynamic" layout is to keep a QList (or other container) of pointers to Widgets and use those pointers to add/remove from the layout.

//you can code add and remove helper functions that add or remove widget pointer both to layout and to QList.
LE:
the remove function also delete the widget pointer.

[lotek]

???
I am always removing the first item cnt times. What's wrong about it?

I've corrected the attached source code, there was the copy & paste bug, the name of the layout was the real one instead of the "aVLayout" used for this example.

---

Added after 4 minutes:

Hello Zlatomir,

This is the possible workaround. But why my code does not work?
And what about empty (without QWidget) QLayout items? Shall they stay ?
Is it a bug or feature?

[Zlatomir]

I don't have time right now to check the QLayoutItem documentation, but you can check it, my guess you shouldn't delete it (or anyway what you want is to delete the contained widget <if any>)

[lotek]

This approach can never work; you're guaranteed to walk off the end of the layout vector once you remove something from it, because you keep looping over the entire initial count before removal.

???
I am always removing the first item cnt times. What's wrong about it?

---

Added after 8 minutes:

I don't have time right now to check the QLayoutItem documentation, but you can check it, my guess you shouldn't delete it (or anyway what you want is to delete the contained widget <if any>)

Unfortunately, there is no word about it.
In the void QLayout::removeItem ( QLayoutItem * item )
it is stated"
"Removes the layout item item from the layout. It is the caller's responsibility to delete the item."
( it is what I attempt do do)

in the description of void QLayout::removeWidget ( QWidget * widget )
there is:
"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.
Note: The ownership of widget remains the same as when it was added."

[DanH]

Zlatomir, there is no interface (that I can find) to remove a widget from a QLayoutItem -- once it's in the layout item you can only delete the layout item.

[To the OP:]
Might we ask why you want to do all this in the first place? Why don't you just delete the entire layout and be done with it?

At the very least, change your loop to work from the end backwards, or just delete item(0) cnt times.

And remember, you need to check the item type. An item may be a spacer, or another layout.

[And thanks to this strange forum software for glomming all my posts together, even though the later ones are directed at the OP.]