How to know if a line intersects a rectangle

Question

I have checked out this question, but the answer is very large for me:

How to know if a line intersects a plane in C#? - Basic 2D geometry

Is there any .NET method to know if a line defined by two points intersects a rectangle?

public bool Intersects(Point a, Point b, Rectangle r)
{
   // return true if the line intersects the rectangle
   // false otherwise
}

Thanks in advance.

Answer 1

    public static bool LineIntersectsRect(Point p1, Point p2, Rectangle r)
    {
        return LineIntersectsLine(p1, p2, new Point(r.X, r.Y), new Point(r.X + r.Width, r.Y)) ||
               LineIntersectsLine(p1, p2, new Point(r.X + r.Width, r.Y), new Point(r.X + r.Width, r.Y + r.Height)) ||
               LineIntersectsLine(p1, p2, new Point(r.X + r.Width, r.Y + r.Height), new Point(r.X, r.Y + r.Height)) ||
               LineIntersectsLine(p1, p2, new Point(r.X, r.Y + r.Height), new Point(r.X, r.Y)) ||
               (r.Contains(p1) && r.Contains(p2));
    }

    private static bool LineIntersectsLine(Point l1p1, Point l1p2, Point l2p1, Point l2p2)
    {
        float q = (l1p1.Y - l2p1.Y) * (l2p2.X - l2p1.X) - (l1p1.X - l2p1.X) * (l2p2.Y - l2p1.Y);
        float d = (l1p2.X - l1p1.X) * (l2p2.Y - l2p1.Y) - (l1p2.Y - l1p1.Y) * (l2p2.X - l2p1.X);

        if( d == 0 )
        {
            return false;
        }

        float r = q / d;

        q = (l1p1.Y - l2p1.Y) * (l1p2.X - l1p1.X) - (l1p1.X - l2p1.X) * (l1p2.Y - l1p1.Y);
        float s = q / d;

        if( r < 0 || r > 1 || s < 0 || s > 1 )
        {
            return false;
        }

        return true;
    }

Answer 2

Unfortunately the wrong answer has been voted up. It is much to expensive to compute the actual intersection points, you only need comparisons. The keyword to look for is "Line Clipping" (http://en.wikipedia.org/wiki/Line_clipping). Wikipedia recommends the Cohen-Sutherland algorithm (http://en.wikipedia.org/wiki/Cohen%E2%80%93Sutherland) when you want fast rejects, which is probably the most common scenario. There is a C++-implementation on the wikipedia page. If you are not interested in actually clipping the line, you can skip most of it. The answer of @Johann looks very similar to that algorithm, but I didn't look at it in detail.