On Thu, Apr 19, 2018 at 6:12 AM, 'Karsten Wolf' via leo-editor <
[email protected]> wrote:
> ...
>
>> Note that apparently simple tests can be fraught. There is a bug against
>> ks.isPlainKey. It ends with:
>>
>> if s in string.printable:
>> return True
>> if len(s) > 1:
>> return False
>> return unicodedata.category(s).startswith('C')
>>
>> Why this fails is mysterious. Perhaps the test should be:
>>
>
> s could still be a str (string.printable is a subset) and make
> unicodedata.category(s) fail.
>
> The question is: Why is s not unicode?
>
An excellent question. Happily, the soon-to-be-pushed version of
ks.isPlainKey bypasses unicodedata entirely:
def isPlainKey(self):
'''
Return True if self.s represents a plain key.
**Note**: The caller is responsible for handling Alt-Ctrl keys.
'''
s = self.s
if s in g.ignoreChars:
# For unit tests.
return False
if self.find_mods(s) or self.isFKey():
return False
if s in self.specialChars:
return False
return True
Once a few special cases are handled, the interesting assumption is that
all keys without modifiers are "plain" keys, even (especially) Chinese
characters. It looks like this is working now.
Edward
--
You received this message because you are subscribed to the Google Groups
"leo-editor" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To post to this group, send email to [email protected].
Visit this group at https://groups.google.com/group/leo-editor.
For more options, visit https://groups.google.com/d/optout.
