function call questions

[Anssi Saari]

"Frank Millman" <frank at chagford.com> writes:

> Let's see if I can explain. I am using 't' and 'r' instead of 'tree'
> and 'root', but otherwise it is the same as your original example.
>
>>>> t = {}
>>>> r = t
>>>> id(t)
> 2542235910088
>>>> id(r)
> 2542235910088
>
> At this point, t and r are both references to the same empty dictionary.
>
>>>> r = r.setdefault('a', {})
>
> This has done two things.
>
> It has inserted the key 'a' into the dictionary, and set its value to {}.
>
>>>> t
> {'a': {}}
>>>> id(t)
> 2542235910088
>
> It has also rebound 'r' so that it now references the new empty
> dictionary that has been inserted.

I guess this is where I fell of the wagon previously. I got it now.

>>>> r
> {}
>>>> id(r)
> 2542234429896
>>>>t['a']
> {}
>>>> id(t['a'])
> 2542234429896
>
> Now continue this process with r = r.setdefault('b', {}), and watch
> what happens.

OK, so what happens is that now t references the dictionary with 
{'a': {}} and r references the empty dict inside that. So when we assign to r
again, it's the empty dict inside t (the one accessed by key 'a') that
changes to {'b': {}} and t becomes {'a': {'b': {}}}.