There is no '1-1-1' case, the case is '1-1-1-pause'. Only after the pause
would the software (may it be kernel or user-space) know that and what
character has been meant. At the '1-1-1' state the user might press again a 1
and then wait, then this might be an '1-pause' case.
The idea is that in the case of '1-1-1-pause' the driver queues exactly one
character, e.g. the "c".
> 2. How do you handle the case where the app is waiting for one key press
> only, but the user hits '1' three times?
See 1. The app doesn't see that this key has been hit three times.
> 3. what if the app doesn't want to accept another character into (say)
> a text field? The effect of "a" "^h" "b" will always replace the last
> character.
When the driver would send Backspaces, then yes. Otherwise not.
So we both agree that sending backspaces from a cell-phone-like keyboard
driver is a bad thing :-)
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