Home Original page

function call questions

Frank Millman frank at chagford.com
Sat Oct 22 05:05:32 EDT 2016 
wrote in message 
news:2853d778-857e-46fc-96a0-8d164c098499 at googlegroups.com...

在 2016年10月20日星期四 UTC+8下午11:04:38，Frank Millman写道：
> wrote in message
> news:01cfd810-0561-40b1-a834-95a73dad6e56 at googlegroups.com...
>
> Hi Frank,
>
> thanks for your kind help. What confused me is at this line:
>
> >>> r = r.setdefault('b', {})
>
> and its previous one
>
> >>> r = r.setdefault('a', {})
>
> When r.setdefault('a',{}) is run, I understand it will return an empty {}. 
> At this time both r & t reference to {'a':{}}, right? So when "r = 
> r.setdefault('a',{})" is run, r reference to {} while t keeps the same as 
> {'a':{}}.
>
> then comes r.setdefault('b',{}). What hinder me is here. since r has 
> changed its reference to {}, r.setdefault('b',{}) will return {} again. So 
> what does this done to t? why at this time t changes to {'a':'b':{}}? 
> Sorry for my silly here. Thanks
>

I don't know if you have been following the other posts in this thread.

There were some posts from Anssi Saari who was also confused by this, but 
then yesterday he sent a post saying that he has now 'got it'.

I don't think I can do better than quote his explanation of what is 
happening -

"""
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': {}}}.
"""

That is exactly right.

Does that help?

Frank
More information about the Python-list mailing list