Szymon <szymon1...@gmail.com> added the comment:
One more thing: It's worth mentioning that by default, you'll get only 28
significant digits of precision, so doing
f"{Decimal('1234567890123456789012.12345678').normalize():f}" will produce
'1234567890123456789012.123457' (two last digits got rounded).
----------
_______________________________________
Python tracker <rep...@bugs.python.org>
<https://bugs.python.org/issue43315>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
