setdefault behaviour question


P

pete McEvoy

I am confused by some of the dictionary setdefault behaviour, I think
I am probably missing the obvious here.

def someOtherFunct():
print "in someOtherFunct"
return 42

def someFunct():
myDict = {1: 2}
if myDict.has_key(1):
print "myDict has key 1"
x = myDict.setdefault(1, someOtherFunct()) # <<<<< I didn't
expect someOtherFunct to get called here
print "x", x
y = myDict.setdefault(5, someOtherFunct())
print "y", y


+++++++++++++++++++++

if I call someFunct() I get the following output

myDict has key 1
in someOtherFunct
x 2
in someOtherFunct
y 42


For the second use of setdefault I do expect a call as the dictionary
does not the key. Will the function, someOtherFunct, in setdefault
always be called anyway and it is just that the dictionary will not be
updated in any way?
 
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P

pete McEvoy

Ah - I have checked some previous posts (sorry, should
have done this first) and I now can see that the
lazy style evaluation approach would not be good.
I can see the reasons it behaves this way.

many thanks anyway.
 

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