Thanks for your explanation, Mike. So, if there is not a bug, I can pass
the outer rect as clip region to attain my aim, and:
Region rgn = new Region();
(1) //----- actual result: The application will crash here with an
exception here
//------expect result : ?
rgn.setPath(p, null);
(2) //----- actual result: The region is the rect area which encircle
the rotated
rect, not the rotated rect itself
//-------expect result :?
rgn.setPath(p, rgn);
(3) //----- actual result: The region is the rect area which encircle
the rotated
rect, not the rotated rect itself
//------- expect result: The region should be a complicate
area, inclined rect clipped by the original rect
Region clipRgn = new Region(top, bottom, left, right);
mRgn2.setPath(p, clipRgn);
I'll try to use outer rect region as the clip region later, see what's
happen currently. Would you please tell me when this region bug can be
fixed? I need try to check if my project can catch up the schedule,
otherwise, I have to try to calculate this region by ourselves, it would
take more efforts.
BR,
-David
On Fri, Feb 20, 2009 at 10:49 PM, Mike Reed <[email protected]> wrote:
>
> Ah, that's a bug, null should be allowed. I'll see what can be done
> there for the future.
>
> The clip parameter is mean to be a hint to speedup turning the path
> into a region by restricting the result to a clipped subset of the
> path. For your purposes, you can just make a big rectangular region
> for the clip. The bounds of the path or larger.
>
> On Thu, Feb 19, 2009 at 10:22 PM, David Hu <[email protected]> wrote:
> > Thanks for your reply, Mike. I've tried your method, seems still not
> > work yet. The second parameter of Region.setPath (clip) can't be null.
> >
> >
> > If we use null, there will be an exception happen. So I've tried to
> use
> > the region I've just constructed or the original rect region, the area is
> > still the ourter standard rect area, not the inclined rect which rotated
> > from a standard rect. Here is my code tip and possible result:
> >
> > //Calculate region
> > top = 150;
> > bottom = top + bmp.getHeight(); //bmp is a bitmap instance
> > left = 200;
> > right = left + bmp.getWidth();
> > Path p = new Path();
> > p.addRect(left, top, right, bottom, Path.Direction.CCW);
> >
> > // use Matrix to rotate 30 degrees
> > Matrix mtx = new Matrix();
> > mtx.setRotate(30);
> > p.transform(mtx);
> >
> > Region rgn = new Region();
> > (1) //----- The application will crash here with an exception here
> > rgn.setPath(p, null);
> > (2) //----- The region is the rect area which encircle the rotated
> > rect, not the rotated rect itself
> > rgn.setPath(p, rgn);
> > (3) //----- The region is the rect area which encircle the rotated
> > rect, not the rotated rect itself
> > Region clipRgn = new Region(top, bottom, left, right);
> > mRgn2.setPath(p, clipRgn);
> > BTW, I searched in android source code and www.google.com, can't find
> any
> > usage of this API:
> >
> > public boolean setPath(Path path, Region clip)
> >
> > So now, my question is which clip region should I pass or any other way
> in
> > order to attain my aim? Hope I've made my aim clearly.
> >
> > BR,
> > -David
> >
> > On Thu, Feb 19, 2009 at 11:27 PM, Mike Reed <[email protected]> wrote:
> >>
> >> You could possibly un-rotate your touch-point by 30 degrees, and then
> >> just use the rectangle.
> >>
> >> However, you can make complex regions by first constructing a Path,
> >> and then calling region.setPath(...), which converts the path into a
> >> region. Below is pseudo sample code:
> >>
> >> Path p = new Path();
> >> p.addRect(rect); // this is your rect
> >> p.transform(matrix); // construct a matrix and then rotate as you wish
> >> region.setPath(p, null);
> >>
> >> On Thu, Feb 19, 2009 at 5:01 AM, <[email protected]> wrote:
> >> >
> >> > I want to judge whether the touch point(x, y) is in a region or
> >> > not, the region is from a stardard rect by rotating specified degrees,
> >> > from example, rotate 30 degrees. There is a class named Region in
> >> > Android, but as I researched, it just supports standard rect, is there
> >> > any other way to judge whether a point is in an acclivitous rect? How
> >> > to do it?
> >> >
> >> > Br,
> >> > -David
> >> > >
> >> >
> >> >>
> >
>
> >
>
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