Bugs item #1296434, was opened at 2005-09-20 10:11 Message generated for change (Comment added) made by tjreedy You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1296434&group_id=5470
Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Documentation Group: Python 2.4 Status: Open Resolution: None Priority: 5 Submitted By: Alan G (abgrover) Assigned to: Nobody/Anonymous (nobody) Summary: Call by object reference sometimes call by value Initial Comment: The tutorial for 2.4.1, section 4.6 Defining Functions states that formal parameters are introduced into the local symbol table, making all calls call by object reference. The footnote points out that this means that changes to mutable objects will be seen by the caller. This is also illustrated in the example involving calling the list method append. It would be helpful if the example could point out that passing a value such as 1 passes an immutable object (the constant integer value 1), and so it is impossible to write code such as: a = 1 def f(val): val = val + 1 and expect that after the call a == 2, even though val == 2. My experience is that this is a confusing issue for new users, who may not understand that val = val + 1 tosses the object reference value passed, replacing it with a new local object. New users tend to see val as a mutable object, since we just changed the value, didn't we? :) ---------------------------------------------------------------------- >Comment By: Terry J. Reedy (tjreedy) Date: 2005-09-23 18:01 Message: Logged In: YES user_id=593130 I agree that there are problems of beginners misunderstanding Python's object model. However, the proposed fix is not exactly correct. Python *always* calls functions by binding local parameter names to argument objects or lists or dicts thereof. Whenever a name is rebound to a new object, it is *always* unbound from the previous object, as it must be. Mutability is irrelevant. So is localness induced by a function call. Changing locality and mutability, your example is equivalent to a = 1 val = a val = val + 1 a = [1] val = a val = val + [2] # or def f(val): val = val + [2] f(a) *all* of which leave 'a' unchanged, but all of which a beginner might think change 'a'. Perhaps you can suggest a different rewording. ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1296434&group_id=5470 _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
