4 Mar
2021
4 Mar
'21
3:50 a.m.
Thank you, type(o) is sufficient. It is possible to use class properties:
type(o).__name__ 'c'
On Wed, Mar 03, 2021 at 10:03:03PM +0000, Paul Bryan wrote:
Since class is a keyword, this is unlikely. Why is type(o) insufficient?
On Wed, 2021-03-03 at 22:59 +0100, Hans Ginzel wrote:
class c: pass o = c() o.__class__
class(o) File "<stdin>", line 1 class(o) ^ SyntaxError: invalid syntax