Thank you, type(o) is sufficient. It is possible to use class properties:

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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:

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

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