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 <r...@google.com> 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 <vistoda...@gmail.com> 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 <r...@google.com> 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, <vistoda...@gmail.com> 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 android-developers@googlegroups.com To unsubscribe from this group, send email to android-developers-unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/android-developers?hl=en -~----------~----~----~----~------~----~------~--~---