BoundingRec from a line with inclination in Qt - c++

Question

I have redefined my own QGraphicsItem to show a LineString. I redefined this because I need to create my own boundingbox and painter redefined the abstract method.
Now I have this code:
QRectF myQGraphicsLine::boundingRect() const
{
double lx = qMin(myLine->getX(0), myLine->getX(1));
double rx = qMax(myLine->getX(0), myLine->getX(1));
double ty = qMin(-myLine->getY(0), -myLine->getY(1));
double by = qMax(-myLine->getY(0), -myLine->getY(1));
return QRectF(lx-size/2,ty, rx -lx + size, by-ty).;
}
void myQGraphicsLine::paint(QPainter * painter, const QStyleOptionGraphicsItem * option, QWidget * widget)
{
pen.setColor(Qt::red);
QLine line(myLine->getX(0), -myLine->getY(0), myLine->getX(1), -myLine->getY(1));
pen.setWidth(size);
painter->setPen(pen);
painter->setBrush(brush);
painter->drawLine(line);
}
This all work fine, but I have a little problem with the boundingRec.
If the line follows the x- or y-axis I get this result:
And in other position I get this:
and I need this:
Does anyone know any way to rotate the boundinRec? Thanks a lot!

Answer 1

I fixed the problem redefining QGraphicsItem::shape() method.
For use this, I created a QPoligonF with my line shape, in my case I used this function:
void myQGraphicsLine::createSelectionPolygon()
{
QPolygonF nPolygon;
QLineF line(myLine->getX(0), -myLine->getY(0), myLine->getX(1), -myLine->getY(1));
qreal radAngle = line.angle()* M_PI / 180; //This the angle of my line vs X axe
qreal dx = size/2 * sin(radAngle);
qreal dy = size/2 * cos(radAngle);
QPointF offset1 = QPointF(dx, dy);
QPointF offset2 = QPointF(-dx, -dy);
nPolygon << line.p1() + offset1
<< line.p1() + offset2
<< line.p2() + offset2
<< line.p2() + offset1;
selectionPolygon = nPolygon;
update();
}
The paint(),boundingRect() and shape() methods stayed like this:
QRectF myQGraphicsLine::boundingRect() const
{
return selectionPolygon.boundingRect();
}
void myQGraphicsLine::paint(QPainter * painter, const QStyleOptionGraphicsItem * option, QWidget * widget)
{
pen.setColor(Qt::red);
QLineF line(myLine->getX(0), -myLine->getY(0), myLine->getX(1), -myLine->getY(1));
pen.setWidth(size);
painter->setPen(pen);
painter->setBrush(brush);
painter->drawLine(line);
}
QPainterPath myQGraphicsLine::shape() const
{
QPainterPath path;
path.addPolygon(selectionPolygon);
return path;
}
Thanks all for your answers!!

Answer 2

A QRect or QRectf cannot be rotated. Its sides are always parallel to the vertical and horizontal axes. Think of it as a point (x, y) giving the upper left corner of the rectangle plus 'width' and 'height' members. No matter what transformation you perform on the QRect it will always interpret 'width' as a horizontal distance and 'height' as vertical distance.
You can extract the four corners of the QRect as points and then rotate those points, but there will be no way to convert the rotated points back to a QRect unless they just happen to be parallel to the sides of the viewport/window.