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 '__main__.c'>
>>> class(o)
File "<stdin>", line 1
class(o)
^
SyntaxError: invalid syntax
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