Ole-M - 1 month ago 5

C++ Question

I have an application where I place several curves in my scene. I was looking for an easy way to detect if the user pressed on the line.

`boundingRect()`

`intersects()`

`selectionMargin`

`y = ax + b`

`bool GraphApp::pointInPath(QPainterPath path, QPointF pos)`

{

qreal posY = pos.y();

qreal posX = pos.x();

for (int i = 0; i < path.elementCount()-1; ++i) {

if (posX < path.elementAt(i + 1).x && posX > path.elementAt(i).x) {

qreal dy = path.elementAt(i + 1).y - path.elementAt(i).y;

qreal dx = path.elementAt(i + 1).x - path.elementAt(i).x;

qreal a = dy / dx;

qreal b = path.elementAt(i).y - (path.elementAt(i).x * a);

if (selectionMargin == 0.0)

selectionMargin = 0.5;

qreal lowerBound = (a * posX + b) + selectionMargin;

qreal upperBound = (a * posX + b) - selectionMargin;

if (posY < lowerBound && posY > upperBound)

return true;

}

}

return false;

}

So it seems like this function returns false when I send a mousePressEvent from the area coverd by the vertical lines. My first thought is the if-sentence:

`if (posX < path.elementAt(i + 1).x && posX > path.elementAt(i).x)`

Any other ideas for how I can implement this without the if-sentence?

I have also seen other people struggling with finding a nice way to check if a

`QPainterPath`

`boundingRect()`

`intersects()`

EDIT: As far as I know,

`contains()`

`boundingRect()`

Answer

I once needed something similar than you. I needed to test two paths for similarity. Therefore I created a path from a list of points (I hope you don't need a more complex path since this solution would become extremely more difficult for general QPaintingPaths). This path is constructed using a given "tolerance", this is your `selectionMargin`

.

The function returns a QPainterPath which "draws a region around the given polyline". This region can then be filled and would result in the same image as drawing the original polyline using a pen width of `tolerance`

using round cap and round join options.

You can also, and this is what you want to do, check if a given point is contained in this path. Note that `QPainterPath::contains`

checks for a point to lie within the closed region defined by the path. E.g., this closed region is empty for a single line segment and a triangle for two line segments, so this is not what you want if you use `contains`

directly on your path (as I mentioned in the 3rd comment to your question).

```
QPainterPath intersectionTestPath(QList<QPointF> input, qreal tolerance)
{
//will be the result
QPainterPath path;
//during the loop, p1 is the "previous" point, initially the first one
QPointF p1 = input.takeFirst();
//begin with a circle around the start point
path.addEllipse(p1, tolerance, tolerance);
//input now starts with the 2nd point (there was a takeFirst)
foreach(QPointF p2, input)
{
//note: during the algorithm, the pair of points (p1, p2)
// describes the line segments defined by input.
//offset = the distance vector from p1 to p2
QPointF offset = p2 - p1;
//normalize offset to length of tolerance
qreal length = sqrt(offset.x() * offset.x() + offset.y() * offset.y());
offset *= tolerance / length;
//"rotate" the offset vector 90 degrees to the left and right
QPointF leftOffset(-offset.y(), offset.x());
QPointF rightOffset(offset.y(), -offset.x());
//if (p1, p2) goes downwards, then left lies to the left and
//right to the right of the source path segment
QPointF left1 = p1 + leftOffset;
QPointF left2 = p2 + leftOffset;
QPointF right1 = p1 + rightOffset;
QPointF right2 = p2 + rightOffset;
//rectangular connection from p1 to p2
{
QPainterPath p;
p.moveTo(left1);
p.lineTo(left2);
p.lineTo(right2);
p.lineTo(right1);
p.lineTo(left1);
path += p; //add this to the result path
}
//circle around p2
{
QPainterPath p;
p.addEllipse(p2, tolerance, tolerance);
path += p; //add this to the result path
}
p1 = p2;
}
//This does some simplification; you should use this if you call
//path.contains() multiple times on a pre-calculated path, but
//you won't need this if you construct a new path for every call
//to path.contains().
return path.simplified();
}
```

Source (Stackoverflow)

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