I've implemented multitouch on Motorola Milestone and it seems to be
working fine. Moreover, I didn't find it too complicated. See the
source below. It can't guarantee that this is 100% correct but at
least it's straight forward:
@Override
public boolean onTouch(View v, MotionEvent event) {
for (int i=0; i<event.getPointerCount(); i++) {
Log.d("Pointer", "Pointer "+(i+1)+": x="+event.getX(i)+",
y="+event.getY(i));
}
}
I'd be glad to hear if this works for anyone else.
On 26 Dez. 2009, 20:49, rageman <nene...@gmail.com> wrote:
> Trying once more, all replies seems to end up in the void :-(
>
> Unfortunately the design of the multi touch api requires new api calls
> introduced in 2.0.
> In my opinion it would be nicer to have added a index for the move
> event in the same way as on the up/down event.
> This would have leave the api fully backwards compatible.
>
> We would like to support platform 1.5 - 2.x with the same SKU and
> support multi-touch on the 2.x devices and single touch on the other.
> Is this still possible when using function calls introduced in 2.0?
> In C/C++ even objective C you could use weak pointers, is this also
> possible in Java on Android?
>
> Any help would be very appreciated :-).
>
> Cheers,
>
> On Dec 26, 7:33 pm, Dianne Hackborn <hack...@android.com> wrote:
>
>
>
> > There is currently no official sample code, but you can look at the
> > PointerLocation (from Dev Tools) code:
>
> >http://android.git.kernel.org/?p=platform/development.git;a=blob;f=ap...
>
> > On Thu, Dec 24, 2009 at 10:30 PM, pcm2a <reeeye...@gmail.com> wrote:
> > > Are there any multitouch code examples that Google provides for SDK
> > > levels 2.0 or higher? I do searches on here and on google and come up
> > > empty.
>
> > > --
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> > --
> > Dianne Hackborn
> > Android framework engineer
> > hack...@android.com
>
> > Note: please don't send private questions to me, as I don't have time to
> > provide private support, and so won't reply to such e-mails. All such
> > questions should be posted on public forums, where I and others can see and
> > answer them.
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