On 2011-05-19, at 11:25 , Łukasz Langa wrote: > Wiadomość napisana przez Stefan Behnel w dniu 2011-05-19, o godz. 10:37: > >>> But why wouldn't "they" expect `b'de' + 1` to work as well in this case? If a 1-byte bytes is equivalent to an integer, why not an arbitrary one as well? >> >> The result of this must obviously be b"de1". > I hope you're joking. At best, the result should be b"de\x01". Actually, if `b'd'+1` returns `b'e'` an equivalent behavior should be that `b'de'+1` returns `b'df'`.
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