Re: Newbie question - default values of a function
Bulba! wrote: Thanks everyone (esp. Fredrik, http://www.effbot.org/zone/index.htm contains lotsa goodies by him I intend to study), I'm going offline to reboot my brain. ;-) Bulba, with reference to your "don't laugh" comment, I'd just like to say that for a newbie your questions do have a habit of getting to the nub of what Python's about. I can see you're going to like this language. regards Steve -- Steve Holden http://www.holdenweb.com/ Python Web Programming http://pydish.holdenweb.com/ Holden Web LLC +1 703 861 4237 +1 800 494 3119 -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question - default values of a function
Thanks everyone (esp. Fredrik, http://www.effbot.org/zone/index.htm contains lotsa goodies by him I intend to study), I'm going offline to reboot my brain. ;-) -- It's a man's life in a Python Programming Association. -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question - default values of a function
"Bulba!" wrote: > Surely the variable i is a mutable object? nope. > OK, again: > def t(a,i=[0]): > ... i[0]=i[0]+1 > ... return (a, i) > ... t(1) > (1, [1]) t(3) > (3, [2]) t(5) > (5, [3]) > > Yes, it works with the list. But why sharing value between calls > pertains some mutable objects and not the other mutable objects? a variable is not an object. reset your brain, and read this: http://www.effbot.org/zone/python-objects.htm or the longer version: http://starship.python.net/crew/mwh/hacks/objectthink.html -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question - default values of a function
On Wed, 22 Dec 2004 22:29:44 GMT, "Matt Gerrans" <[EMAIL PROTECTED]> wrote: >Actually i was not mutable. Try this: >i = 1 >id(i) >i += 1 >id(i) Looks like I will have to read the Language Reference anyway. :-( >Because you are assigning the local reference variable L to a new list, >instead of modifying that original default list that was created. >Is that more clear, or is it now more unclear? ;) Clear now, but I have to say for someone who's been doing some C programming in the past it is pretty confusing, because I'm used to think of any variable as basically a place in memory with a pointer to it, so anything without #define or const was considered "mutable" (gawd I still hate pointers). C-style thinking considered harmful when programming in Python. :o) -- It's a man's life in a Python Programming Association. -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question - default values of a function
Actually i was not mutable. Try this: i = 1 id(i) i += 1 id(i) Change both of your sample functions to also print the id of the default variable and that might shed a little more light on the matter. "i = i + 1" is only changing the local reference called i to refer to an instance of a different integer object.If you change the original function to something similar, it will behave similarly: def f( a, L=[]): L = L + [a] return L Because you are assigning the local reference variable L to a new list, instead of modifying that original default list that was created. Is that more clear, or is it now more unclear? ;) -- http://mail.python.org/mailman/listinfo/python-list
Re: Newbie question - default values of a function
Bulba! wrote: > [...] > But why the following happens? > > def j(a,i=0): > > ... i=i+1 > ... return (a, i) > ... > j(2) > > (2, 1) > j(3) > > (3, 1) > j(5) > > (5, 1) > > > From Language Reference: > > http://www.python.org/doc/2.3.4/ref/function.html > > "Default parameter values are evaluated when the function definition > is executed. This means that the expression is evaluated once, when > the function is defined, and that that same ``pre-computed'' value is > used for each call. This is especially important to understand when a > default parameter is a mutable object, such as a list or a dictionary: > if the function modifies the object (e.g. by appending an item to a > list), the default value is in effect modified. " > > > Surely the variable i is a mutable object? > [...] No, i is not mutable! i is a name for a number object which is _not_ mutable (http://www.python.org/doc/2.4/ref/types.html). The line i=i+1 creates a new number object with the value 1 and gives it the name i. Bye, Dennis -- http://mail.python.org/mailman/listinfo/python-list
Newbie question - default values of a function
OK. Don't laugh. There's this example from tutorial, showing how default value is computed once when defining function and shared between function calls: --- def f(a, L=[]): L.append(a) return L print f(1) print f(2) print f(3) This will print [1] [1, 2] [1, 2, 3] --- (PythonWin 2.3.4 (#53, Oct 18 2004, 20:35:07) [MSC v.1200 32 bit (Intel)] on win32.) Tried it, it works as expected. But why the following happens? >>> def j(a,i=0): ... i=i+1 ... return (a, i) ... >>> >>> j(2) (2, 1) >>> j(3) (3, 1) >>> j(5) (5, 1) >From Language Reference: http://www.python.org/doc/2.3.4/ref/function.html "Default parameter values are evaluated when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that that same ``pre-computed'' value is used for each call. This is especially important to understand when a default parameter is a mutable object, such as a list or a dictionary: if the function modifies the object (e.g. by appending an item to a list), the default value is in effect modified. " Surely the variable i is a mutable object? OK, again: >>> def t(a,i=[0]): ... i[0]=i[0]+1 ... return (a, i) ... >>> t(1) (1, [1]) >>> t(3) (3, [2]) >>> t(5) (5, [3]) Yes, it works with the list. But why sharing value between calls pertains some mutable objects and not the other mutable objects? I'm stymied. -- It's a man's life in a Python Programming Association. -- http://mail.python.org/mailman/listinfo/python-list
