[Tutor] How is "set(ls).add('a') evaluated? [Was: Re: A program that can check if all elements of the list are mutually disjoint]

[boB Stepp]

As is often the case Mr. Singh's query got me to play around in the interpreter.

On Sat, Jun 5, 2021 at 8:10 AM Manprit Singh <manpritsinghece at gmail.com> wrote:

> ls = ["amba", "Joy", "Preet"]
> for idx, ele in enumerate(ls):
>     if idx < len(ls)-1:
>         if not all(set(ele).isdisjoint(ch)for ch in ls[idx+1:]):
>             print("Not mutually disjoint")
>             break
[...]

I have not played around with set's methods and operators to date, so
while trying to understand this code I tried out different things in
the interpreter.  Along the way I tried something and it surprised me:

Python 3.9.5 (tags/v3.9.5:0a7dcbd, May  3 2021, 17:27:52) [MSC v.1928
64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> ls = ["amba", "Joy", "Preet"]
>>> z = set(ls).add('a')
>>> z
>>> print(z)
None           # This surprised me.  I was expecting {'amba', 'Joy',
'Preet', 'a'}.

The "normal" way of using the add() method works fine:

>>> z = set(ls)
>>> z
{'amba', 'Joy', 'Preet'}
>>> z.add('a')
>>> z
{'amba', 'Joy', 'Preet', 'a'}

But the following works which is in a similar chained format:

>>> zz = set(ls).union('a')
>>> zz
{'amba', 'Joy', 'Preet', 'a'}   # BTW, please forgive bad identifier naming!

So I am *not* understanding how "set(ls).add('a')" is evaluated to
result in "None".

TIA!
boB Stepp