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在 2016年10月20日星期四 UTC+8下午11:04:38，Frank Millman写道：
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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
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