pop() clarification

Christophe chris.cavalaria at free.fr
Thu Apr 12 10:29:33 CEST 2007


Carl Banks a écrit :
> On Apr 11, 3:10 pm, "7stud" <bbxx789_0... at yahoo.com> wrote:
>> On Apr 11, 10:44 am, "Scott" <s_brosci... at comcast.net> wrote:
>>
>>
>>
>>> As said before I'm new to programming, and I need in depth explaination to
>>> understand everything the way I want to know it, call it a personality quirk
>>> ;p.
>>> With pop() you remove the last element of a list and return its value:
>>> Now I know list is a bad name, but for the sake of arguement lets assume its
>>> not a built in sequence>
>>>>>> list = ['this', 'is', 'an', 'example']
>>>>>> list.pop()
>>> 'example'
>>>>>> list
>>> ['this', 'is', 'an']
>>> I understand all that.  What I don't understand is why all the documentation
>>> I see says, "When removing a specific element from a list using pop() it
>>> must be in this format: list.pop([i]).
>>> At first I took that to mean that list.pop(i) would return some type of
>>> error, but it doesn't.
>> It's understandable that the definition of pop() is confusing in that
>> way.  It looks like the argument should be a list.  As others have
>> said, that is not what the brackets mean when the documents show the
>> formal definition of a function.
> 
> I wonder if the documentation could take advantage of Python 3000
> annotation syntax.  So
> 
> pop([x])
> 
> would be replaced in the docs by
> 
> pop(x: OPTIONAL)
> 
> Just a thought, probably not a good one.  The brackets are so
> pervasive that it's probably better to just let newbies be confused
> for a little bit.

I'd rather go for the overloading syntax. ie:


pop(i)
pop()

Remove the item at the given position in the list, and return it. If no 
index is specified, a.pop() removes and returns the last item in the list.



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