a problem about "print"

levi nie levinie001 at gmail.com
Wed Jul 4 10:21:27 CEST 2012


Yes,you're right.
I got it now.I mistaked that aList also had be changed in code1 before.
Thanks a lot.

2012/7/4 Matteo Boscolo <matteo.boscolo at boscolini.eu>

>  in the code2
>
> aList=[1,2,3,4,5,6,7,8,9,10]
> aList=str(aList) #<--- here you convert the list in a string
>
>  print aList
> print aList[2] #<-- here you are printing the third caracter of the string
> '[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]'  not the list '[1, 2, 3, 4, 5, 6, 7, 8,
> 9, 10]'
>
> regards
> Matteo
>
>
>
> Il 04/07/2012 09:28, levi nie ha scritto:
>
> Hi,Harrison.
> Your method is cool.
> But i doubt this, if bList and aList just are attached to the same List
> when i write bList=aList,but why the output of the following two code are
> different?
>
>  code1:
>  aList=[1,2,3,4,5,6,7,8,9,10]
> bList=aList
> bList=str(bList)
> print aList
> print aList[2]
>
>  code2:
>  aList=[1,2,3,4,5,6,7,8,9,10]
> aList=str(aList)
> print aList
> print aList[2]
>
>  i'm puzzled now.
>
> 2012/7/4 Harrison Morgan <harrison.morgan at gmail.com>
>
>>
>>
>> On Wed, Jul 4, 2012 at 12:38 AM, levi nie <levinie001 at gmail.com> wrote:
>>
>>> that's good,thanks.
>>> new problem.
>>> when i write
>>> bList=aList
>>> del bList[2]
>>> bList and aList both change,how can i make aList not changed?
>>>
>>>>
>>>
>>  Lists are mutable. That means that when you do bList = aList, you're
>> just creating another reference to aList. They both point to the same list,
>> just using different names. You should read up a bit on immutable vs.
>> mutable objects. Here's something that I found that might explain it a bit
>> better. http://henry.precheur.org/python/copy_list
>>
>
>
>
>
>
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>
>
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