Removing widgets from QGridLayout

2019-01-15 07:58发布

I try to remove widgets from a specified row in a QGridLayout like this:

void delete_grid_row(QGridLayout *layout, int row)
{
    if (!layout || row < 0) return;

    for (int i = 0; i < layout->columnCount(); ++i) {
        QLayoutItem* item = layout->itemAtPosition(row, i);
        if (!item) continue;

        if (item->widget()) {
            layout->removeWidget(item->widget());
        } else {
            layout->removeItem(item);
        }
        delete item;
    }
}

But when I call it, the app crashes with SIGSEGV on delete item in the first iteration. Any ideas?

2条回答
Summer. ? 凉城
2楼-- · 2019-01-15 08:33

The QGridLayout itself is managing the QLayoutItem's. I believe the moment you call removeWidget the item will be deleted. Thus you have an invalid pointer at that point. Attempting to do anything with it, not just delete, will fail.

Thus, just don't delete it, you'll be fine.

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看我几分像从前
3楼-- · 2019-01-15 08:52

Short answer: Use the code provided below

Removing a row or column from a QGridLayout is tricky. Use the removeRow() and removeColumn() functions from the code provided below.

Long answer: Digging into QGridLayout details

First, note that QGridLayout::rowCount() and QGridLayout::columnCount() always return the number of internally allocated rows and columns in the grid layout. As an example, if you call QGridLayout::addWidget(widget,5,7) on a freshly constructed grid layout, the row count will be 6 and the column count will be 8, and all cells of the grid layout except the cell on index (5,7) will be empty and thus invisible within the GUI.

Note that it's unfortunately impossible to remove such an internal row or column from the grid layout. In other words, the row and column count of a grid layout can always only grow, but never shrink.

What you can do is to remove the cell contents of a row or column, which will effectively have the same visual effect as removing the row or column itself. But this of course means that all row and column counts and indices will remain unchanged.

So how can the contents of a row or column be cleared? This unfortunately also isn't as easy as it might seem.

First, you need to think about if you only want to remove the widgets from the layout, or if you also want them to become deleted. If you only remove the widgets from the layout, you must put them back into a different layout afterwards or manually give them a reasonable geometry. If the widgets also become deleted, they will disappear from the GUI. The provided code uses a boolean parameter to control widget deletion.

Next, you have to consider that a layout cell can not just only contain a widget, but also a nested layout, which itself can contain nested layouts, and so on. You further need to handle layout items which span over multiple rows and columns. And, finally, there are some row and column attributes like minimum widths and heights which don't depend on the actual cell contents but still have to be taken care of.

The code

Use removeRow() and removeColumn() to remove a row or column from a given QGridLayout.

/**
 * Removes all layout items on the given @a row from the given grid
 * @a layout. If @a deleteWidgets is true, all concerned child widgets
 * become not only removed from the layout, but also deleted. Note that
 * this function doesn't actually remove the row itself from the grid
 * layout, as this isn't possible (i.e. the rowCount() and row indices
 * will stay the same after this function has been called).
 */
void removeRow(QGridLayout *layout, int row, bool deleteWidgets) {
    remove(layout, row, -1, deleteWidgets);
    layout->setRowMinimumHeight(row, 0);
    layout->setRowStretch(row, 0);
}

/**
 * Removes all layout items on the given @a column from the given grid
 * @a layout. If @a deleteWidgets is true, all concerned child widgets
 * become not only removed from the layout, but also deleted. Note that
 * this function doesn't actually remove the column itself from the grid
 * layout, as this isn't possible (i.e. the columnCount() and column
 * indices will stay the same after this function has been called).
 */
void removeColumn(QGridLayout *layout, int column, bool deleteWidgets) {
    remove(layout, -1, column, deleteWidgets);
    layout->setColumnMinimumWidth(column, 0);
    layout->setColumnStretch(column, 0);
}

/**
 * Helper function. Removes all layout items within the given @a layout
 * which either span the given @a row or @a column. If @a deleteWidgets
 * is true, all concerned child widgets become not only removed from the
 * layout, but also deleted.
 */
void remove(QGridLayout *layout, int row, int column, bool deleteWidgets) {
    // We avoid usage of QGridLayout::itemAtPosition() here to improve performance.
    for (int i = layout->count() - 1; i >= 0; i--) {
        int r, c, rs, cs;
        layout->getItemPosition(i, &r, &c, &rs, &cs);
        if ((r <= row && r + rs - 1 >= row) || (c <= column && c + cs - 1 >= column)) {
            // This layout item is subject to deletion.
            QLayoutItem *item = layout->takeAt(i);
            if (deleteWidgets) {
                deleteChildWidgets(item);
            }
            delete item;
        }
    }
}

/**
 * Helper function. Deletes all child widgets of the given layout @a item.
 */
void deleteChildWidgets(QLayoutItem *item) {
    if (item->layout()) {
        // Process all child items recursively.
        for (int i = 0; i < item->layout()->count(); i++) {
            deleteChildWidgets(item->layout()->itemAt(i));
        }
    }
    delete item->widget();
}
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