Hi, Mike
It still doesn't work for me. I've tried two use case:
1) Pass a very large rect as the clip region------the actual result
is the outer rect, bigger than my expect region;
2) Pass the outer rect as the clip region-----------the actual result
is smaller than the rotated bitmap region, my expect region.
Is there still with some bugs of Region or Path?
My expect use case, I have a ImageView with a Bitmap, after rotate 30
degrees, the bitmap region is a rotated rect from original rect, and the
ImageView containing the bitmap is a bigger rect than the original rect. I
want to judge whether the touch point is in the bitmap region. It seems easy
to judge in the ImageView region, but I want to get the exact bitmap region
for exact judgement. That's the reason why I need use Region and Path.
Br.
-David
On Sat, Feb 21, 2009 at 5:01 AM, Mike Reed <[email protected]> wrote:
>
> By bug, I mean I would like to relax the restriction, and allow you to
> pass null. I don't know yet when the could get in. Thus you should
> always (for now) pass in a region as the clip. It can be something
> large with no downside (i.e. -10,000, .... 10,000)
>
> On Fri, Feb 20, 2009 at 3:57 PM, David Hu <[email protected]> wrote:
> > 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
> >> >> > >
> >> >> >
> >> >> >>
> >> >
> >> >>
> >
>
> >
>
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google
Groups "Android Developers" group.
To post to this group, send email to [email protected]
To unsubscribe from this group, send email to
[email protected]
For more options, visit this group at
http://groups.google.com/group/android-developers?hl=en
-~----------~----~----~----~------~----~------~--~---
