setdefault behaviour question

Discussion in 'Python' started by pete McEvoy, May 19, 2012.

  1. pete McEvoy

    pete McEvoy Guest

    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?
     
    pete McEvoy, May 19, 2012
    #1
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  2. pete McEvoy

    pete McEvoy Guest

    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.
     
    pete McEvoy, May 19, 2012
    #2
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