Re: a problem about "print"
 

@Matteo, @Levi, please don't top post it makes following a thread very
difficult, no other comments, TIA.

On 04/07/2012 08:51, Matteo Boscolo wrote:
> 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@gmail.com
>> <mailto:harrison.morgan@gmail.com>>
>>
>>
>>
>> On Wed, Jul 4, 2012 at 12:38 AM, levi nie <levinie001@gmail.com
>> <mailto:levinie001@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
>>
>>
>>
>>
>
>
>
>
>


--
Cheers.

Mark Lawrence.