On 29/06/2012, at 1:17 PM, Quincey Morris wrote: > Well, this looks wrong to me. In general, each of the 2 touches is going to > begin at a different time, which means this method will be called twice, and > there's only be one touch in the set at each call. (The touch that began > earlier is in a different phase at this event.) > Hmm, ok, I can investigate that. I imagined that touches that went down very close in time were coalesced by the driver into a single event. That assumption, possibly faulty, was strongly reinforced by the example code listed in the documentation, which specifically does exactly this - [touches count] == 2.
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