Yes. In order to call D.get( ) it needs to pass two arguments. The first is
'a', simple. The second is the result of a call to get_default(). So, that
is called. From INSIDE get_default() it prints 'Nobody expects this' but you
should expect it, get_default() gets executed. Following that it prints '1',
because the default value was NOT USED. If it was used, you would see 'Nobody
expects this' followed by 0.
--- Joseph S.
-----Original Message-----
From: Mark Polesky <markpole...@yahoo.com>
Sent: Tuesday, December 15, 2020 12:07 PM
To: python-list@python.org
Subject: dict.get(key, default) evaluates default even if key exists
Hi.
# Running this script....
D = {'a':1}
def get_default():
print('Nobody expects this')
return 0
print(D.get('a', get_default()))
# ...generates this output:
Nobody expects this
1
###
Since I'm brand new to this community, I thought I'd ask here first... Is this
worthy of a bug report? This behavior is definitely unexpected to me, and I
accidentally coded an endless loop in a mutual recursion situation because of
it. Calling dict.get.__doc__ only gives this short sentence: Return the value
for key if key is in the dictionary, else default. Nothing in that docstring
suggests that the default value is evaluated even if the key exists, and I
can't think of any good reason to do so.
Am I missing something?
Thanks,
Mark
--
https://mail.python.org/mailman/listinfo/python-list
