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 +, 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:
>>>
Err, why is it insufficient?
On Wed, 2021-03-03 at 14:03 -0800, Paul Bryan wrote:
> Since class is a keyword, this is unlikely. Why is type(o) not
> insufficient?
>
> On Wed, 2021-03-03 at 22:59 +0100, Hans Ginzel wrote:
> > > > > class c: pass
> > > > > o = c()
> > > > > o.__class__
> >
> > >
Since class is a keyword, this is unlikely. Why is type(o) not
insufficient?
On Wed, 2021-03-03 at 22:59 +0100, Hans Ginzel wrote:
> > > > class c: pass
> > > > o = c()
> > > > o.__class__
>
> > > > class(o)
> File "", line 1
> class(o)
> ^
> SyntaxError: invalid syntax
On Tue, Feb 9, 2021 at 6:39 PM Hans Ginzel wrote:
>
> Please, consider class(obj) to return obj.__class__
> consistenly with dir(), vars(), repr(), str(),…
>
> >>> class c: pass
> >>> o = c()
> >>> o.__class__
>
> >>> class(o)
>File "", line 1
> class(o)
> ^
> SyntaxError:
Please use type(o) instead.
On Tue, Feb 9, 2021 at 4:36 PM Hans Ginzel wrote:
>
> Please, consider class(obj) to return obj.__class__
> consistenly with dir(), vars(), repr(), str(),…
>
> >>> class c: pass
> >>> o = c()
> >>> o.__class__
>
> >>> class(o)
>File "", line 1
> class(o)
>
