QGridLayout: Getting the list of QWidget added

Suppose I have something like:

Code:

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QGridLayout layout;
layout.addWidget(new QWidget());
layout.addWidget(new QWidget());

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What is the method that I can use to get the list of the added QWidgets ?

Something like the imaginary "getAddedWidgets()" below:

Code:

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QList<QWidget*> addedWidgets = layout.getAddedWidgets(); 
Q_ASSERT( addedWidgets.size() == 2 );

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Added after 1 29 minutes:

The example code below shows how you could iterate over each item:

Code:

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QGridLayout layout;
 
      // add 3 items to layout
      layout.addItem(new QSpacerItem(2,3), 0, 0, 1, 1);
      layout.addWidget(new QWidget);
      layout.addWidget(new QWidget);
 
      // sanity checks
      Q_ASSERT(layout.count() == 3);
      Q_ASSERT(layout.itemAt(0));
      Q_ASSERT(layout.itemAt(1));
      Q_ASSERT(layout.itemAt(2));
      Q_ASSERT(layout.itemAt(3) == NULL);
 
      // iterate over each, only looking for QWidgetItem
      for(int idx = 0; idx < layout.count(); idx++)
      {
        QLayoutItem * item = layout.itemAt(idx);
        if(dynamic_cast<QWidgetItem *>(item))       <-- Note! QWidgetItem, and not QWidget!
          item->widget()->hide();                   <-- widget() will cast you a QWidget!
      }

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dynamic_cast should be avoided when dealing with QObjects.
Use qobject_cast instead.

Quote:

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Originally Posted by tbscope

dynamic_cast should be avoided when dealing with QObjects.
Use qobject_cast instead.

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why so? is dynamic_cast is slower than qobject_cast? I do have RTTI always in my compiler, so given that, can you plz elaborate?