Python primer - comments appreciated!

Terry Reedy tjreedy at udel.edu
Wed Sep 11 16:35:44 CEST 2002


"Thorsten Kampe" <thorsten at thorstenkampe.de> wrote in message
news:alnipo$1pqmu8$1 at ID-77524.news.dfncis.de...
> *  Terry Reedy
> > Then
> > append to the list.  This working depends on there being *two*
> > references to the list.  The one in the dict (possibly new,
possible
> > not) and the one returned by the method and used by append().

Please reread second sentence!!!

> Okay, let's make this clear:
> D = {'0': [3, 6], '1': [4]}
> D.setdefault('1', [])
>   ...returns: [4]

It returns a reference to the list with current value [4]

> [4].append(7)
>  ...puts [4, 7] on the "stack"

NOOOOO!!!!!  In context of of original expression, it appends 7 to
list pointed to by *both* dict and reference returned by setdefault().

> But where is the instruction saying "store [4, 7] back to D['1']"?

The list is never taken out of D['1'].  The append is 'in place' via
the double reference!
Try
print id(D['1']), id(D.setdefault('1', []))
to see that same list.

> Shouldn't this be: D['1'] = D.setdefault('1', []).append(7) ?

Since append returns None, this will not work.

Terry J. Reedy






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