Re: initialized list: strange behavior
Jason Scheirer [EMAIL PROTECTED] writes: On Nov 24, 10:34 pm, [EMAIL PROTECTED] wrote: Hi Python experts! Please explain this behavior: nn=3*[[]] nn [[], [], []] mm=[[],[],[]] mm [[], [], []] Up till now, 'mm' and 'nn' look the same, right? Nope! mm[1].append(17) mm [[], [17], []] nn[1].append(17) nn [[17], [17], [17]] ??? Python 2.5 Win XP Thanks! You're creating three references to the same list with the multiplication operator. There's no need to introduce references: you're creating a list with the same object at each position. [...] Python is pass-by-reference, not pass-by-value. It's certainly not pass-by-reference, nor is it pass-by-value IMHO. -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: initialized list: strange behavior
On Tue, Nov 25, 2008 at 9:23 AM, Arnaud Delobelle [EMAIL PROTECTED]wrote: Jason Scheirer [EMAIL PROTECTED] writes: On Nov 24, 10:34 pm, [EMAIL PROTECTED] wrote: Hi Python experts! Please explain this behavior: nn=3*[[]] nn [[], [], []] mm=[[],[],[]] mm [[], [], []] Up till now, 'mm' and 'nn' look the same, right? Nope! mm[1].append(17) mm [[], [17], []] nn[1].append(17) nn [[17], [17], [17]] ??? Python 2.5 Win XP Thanks! You're creating three references to the same list with the multiplication operator. There's no need to introduce references: you're creating a list with the same object at each position. [...] Python is pass-by-reference, not pass-by-value. It's certainly not pass-by-reference, nor is it pass-by-value IMHO. Please don't get into this here. We have enough threads for this already. -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: initialized list: strange behavior
Arnaud Delobelle wrote: Jason Scheirer [EMAIL PROTECTED] writes: On Nov 24, 10:34 pm, [EMAIL PROTECTED] wrote: Hi Python experts! Please explain this behavior: nn=3*[[]] nn [[], [], []] mm=[[],[],[]] mm [[], [], []] Up till now, 'mm' and 'nn' look the same, right? Nope! mm[1].append(17) mm [[], [17], []] nn[1].append(17) nn [[17], [17], [17]] ??? Python 2.5 Win XP Thanks! You're creating three references to the same list with the multiplication operator. There's no need to introduce references: you're creating a list with the same object at each position. [...] Python is pass-by-reference, not pass-by-value. It's certainly not pass-by-reference, nor is it pass-by-value IMHO. Since no lists are being passed as arguments in these examples it's not pass-by-anything. Jump off that horse right now! regards Steve -- Steve Holden+1 571 484 6266 +1 800 494 3119 Holden Web LLC http://www.holdenweb.com/ -- http://mail.python.org/mailman/listinfo/python-list
Re: initialized list: strange behavior
The issue is exhausted in Python Library Reference, Chapter 3.6, so I should apologize for initial posting. All comments were helpful, though Arnaud and Steve are right that pass-by-anything is off the point. Thanks All! -- http://mail.python.org/mailman/listinfo/python-list
Re: initialized list: strange behavior
Steve Holden [EMAIL PROTECTED] writes: Arnaud Delobelle wrote: Jason Scheirer [EMAIL PROTECTED] writes: Python is pass-by-reference, not pass-by-value. It's certainly not pass-by-reference, nor is it pass-by-value IMHO. Since no lists are being passed as arguments in these examples it's not pass-by-anything. Jump off that horse right now! You're right that Python's calling semantics have nothing to do with the question asked. I just thought best not to leave the OP under any misapprehension. -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
initialized list: strange behavior
Hi Python experts! Please explain this behavior: nn=3*[[]] nn [[], [], []] mm=[[],[],[]] mm [[], [], []] Up till now, 'mm' and 'nn' look the same, right? Nope! mm[1].append(17) mm [[], [17], []] nn[1].append(17) nn [[17], [17], [17]] ??? Python 2.5 Win XP Thanks! -- http://mail.python.org/mailman/listinfo/python-list
Re: initialized list: strange behavior
[EMAIL PROTECTED] wrote: Hi Python experts! Please explain this behavior: [] make an empty list. [ [],[],[] ] makes a list of three different empty lists. 3*[[]] makes a list of three references to the same list. Realy, that should explain it all, but perhaps there are enough empty lists here to obscure things. Try this: Here b is a list that contains three references to a. Modify a, and all three references to a show the modification: a = [1,2,3] b = [a,a,a] a.append(4) b [[1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]] Gary Herron nn=3*[[]] nn [[], [], []] mm=[[],[],[]] mm [[], [], []] Up till now, 'mm' and 'nn' look the same, right? Nope! mm[1].append(17) mm [[], [17], []] nn[1].append(17) nn [[17], [17], [17]] ??? Python 2.5 Win XP Thanks! -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: initialized list: strange behavior
On Nov 24, 10:34 pm, [EMAIL PROTECTED] wrote: Hi Python experts! Please explain this behavior: nn=3*[[]] nn [[], [], []] mm=[[],[],[]] mm [[], [], []] Up till now, 'mm' and 'nn' look the same, right? Nope! mm[1].append(17) mm [[], [17], []] nn[1].append(17) nn [[17], [17], [17]] ??? Python 2.5 Win XP Thanks! You're creating three references to the same list with the multiplication operator. You can easily get the same behavior because of similar mechanics in a more common scenario: In [1]: a = [] In [2]: b = a In [3]: a, b Out[3]: ([], []) In [4]: a.append(100) In [5]: a, b Out[5]: ([100], [100]) Python is pass-by-reference, not pass-by-value. -- http://mail.python.org/mailman/listinfo/python-list
