Re: [Tutor] Functions returning multiple values
I have an update: I can easily undertand why this example doesn't work: def nochange(x): x = 0 y = 1 nochange(y) print y # Prints out 1 X is a local variable, and only gets modified in the function, that doesn't return any value. But it's very difficult for me to understand WHY this works: def change(some_list): some_list[1] = 4 x = [1,2,3] change(x) print x # Prints out [1,4,3] some_list is a local list, isn't it? Maybe i can't have lists that are only existing in a function? Thankyou all 2010/2/22 Kent Johnson ken...@tds.net On Mon, Feb 22, 2010 at 9:13 AM, Giorgio anothernetfel...@gmail.com wrote: And, i have some difficulties understanding the other strange example in that howto. Just scroll down to: However, the point is that the value of x is picked up from the environment at the time when the function is defined. How is this useful? Let’s take an example — a function which composes two other functions. He is working on a function that compose other 2 functions. This is the final solution def compose(fun1, fun2): def inner(x, fun1=fun1, fun2=fun2): return fun1(fun2(x)) return inner But also tries to explain why this example: # Wrong version def compose(fun1, fun2): def inner(x): return fun1(fun2(x)) return inner def fun1(x): return x + world! def fun2(x): return Hello, sincos = compose(sin,cos) # Using the wrong version x = sincos(3) Won't work. Now, the problem is that the inner function gets fun1 and fun2 from other 2 functions. My question is: why? inner is a sub-function of compose, where fun1 and fun2 are defined. It does work: In [6]: def compose(fun1, fun2): ...: def inner(x): ...: return fun1(fun2(x)) ...: return inner ...: In [7]: def fun1(x): ...: return x + world! ...: In [8]: def fun2(x): ...: return Hello, ...: In [9]: from math import sin, cos In [10]: sincos = compose(sin,cos) # Using the wrong version In [11]: In [12]: x = sincos(3) In [13]: In [14]: x Out[14]: -0.8360218615377305 That is a very old example, from python 2.1 or before where nested scopes were not supported. See the note A Note About Python 2.1 and Nested Scopes - that is now the default behaviour. Kent -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
Giorgio wrote: I have an update: I can easily undertand why this example doesn't work: def nochange(x): x = 0 y = 1 nochange(y) print y # Prints out 1 X is a local variable, and only gets modified in the function, that doesn't return any value. But it's very difficult for me to understand WHY this works: def change(some_list): some_list[1] = 4 x = [1,2,3] change(x) print x # Prints out [1,4,3] some_list is a local list, isn't it? Maybe i can't have lists that are only existing in a function? Thankyou all 2010/2/22 Kent Johnson ken...@tds.net mailto:ken...@tds.net On Mon, Feb 22, 2010 at 9:13 AM, Giorgio anothernetfel...@gmail.com mailto:anothernetfel...@gmail.com wrote: And, i have some difficulties understanding the other strange example in that howto. Just scroll down to: However, the point is that the value of x is picked up from the environment at the time when the function is defined. How is this useful? Let’s take an example — a function which composes two other functions. He is working on a function that compose other 2 functions. This is the final solution def compose(fun1, fun2): def inner(x, fun1=fun1, fun2=fun2): return fun1(fun2(x)) return inner But also tries to explain why this example: # Wrong version def compose(fun1, fun2): def inner(x): return fun1(fun2(x)) return inner def fun1(x): return x + world! def fun2(x): return Hello, sincos = compose(sin,cos) # Using the wrong version x = sincos(3) Won't work. Now, the problem is that the inner function gets fun1 and fun2 from other 2 functions. My question is: why? inner is a sub-function of compose, where fun1 and fun2 are defined. It does work: In [6]: def compose(fun1, fun2): ...: def inner(x): ...: return fun1(fun2(x)) ...: return inner ...: In [7]: def fun1(x): ...: return x + world! ...: In [8]: def fun2(x): ...: return Hello, ...: In [9]: from math import sin, cos In [10]: sincos = compose(sin,cos) # Using the wrong version In [11]: In [12]: x = sincos(3) In [13]: In [14]: x Out[14]: -0.8360218615377305 That is a very old example, from python 2.1 or before where nested scopes were not supported. See the note A Note About Python 2.1 and Nested Scopes - that is now the default behaviour. Kent -- -- AnotherNetFellow Email: anothernetfel...@gmail.com mailto:anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor Take a look at the Python gothcha's: http://www.ferg.org/projects/python_gotchas.html#contents_item_6 -- Kind Regards, Christian Witts ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
Thankyou. It's clear, this definitely helps me with the first question. But still doesn't explain almost anything about the example i've posted in the last post, right? Giorgio 2010/2/23 Christian Witts cwi...@compuscan.co.za Giorgio wrote: I have an update: I can easily undertand why this example doesn't work: def nochange(x): x = 0 y = 1 nochange(y) print y # Prints out 1 X is a local variable, and only gets modified in the function, that doesn't return any value. But it's very difficult for me to understand WHY this works: def change(some_list): some_list[1] = 4 x = [1,2,3] change(x) print x # Prints out [1,4,3] some_list is a local list, isn't it? Maybe i can't have lists that are only existing in a function? Thankyou all 2010/2/22 Kent Johnson ken...@tds.net mailto:ken...@tds.net On Mon, Feb 22, 2010 at 9:13 AM, Giorgio anothernetfel...@gmail.com mailto:anothernetfel...@gmail.com wrote: And, i have some difficulties understanding the other strange example in that howto. Just scroll down to: However, the point is that the value of x is picked up from the environment at the time when the function is defined. How is this useful? Let’s take an example — a function which composes two other functions. He is working on a function that compose other 2 functions. This is the final solution def compose(fun1, fun2): def inner(x, fun1=fun1, fun2=fun2): return fun1(fun2(x)) return inner But also tries to explain why this example: # Wrong version def compose(fun1, fun2): def inner(x): return fun1(fun2(x)) return inner def fun1(x): return x + world! def fun2(x): return Hello, sincos = compose(sin,cos) # Using the wrong version x = sincos(3) Won't work. Now, the problem is that the inner function gets fun1 and fun2 from other 2 functions. My question is: why? inner is a sub-function of compose, where fun1 and fun2 are defined. It does work: In [6]: def compose(fun1, fun2): ...: def inner(x): ...: return fun1(fun2(x)) ...: return inner ...: In [7]: def fun1(x): ...: return x + world! ...: In [8]: def fun2(x): ...: return Hello, ...: In [9]: from math import sin, cos In [10]: sincos = compose(sin,cos) # Using the wrong version In [11]: In [12]: x = sincos(3) In [13]: In [14]: x Out[14]: -0.8360218615377305 That is a very old example, from python 2.1 or before where nested scopes were not supported. See the note A Note About Python 2.1 and Nested Scopes - that is now the default behaviour. Kent -- -- AnotherNetFellow Email: anothernetfel...@gmail.com mailto:anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor Take a look at the Python gothcha's: http://www.ferg.org/projects/python_gotchas.html#contents_item_6 -- Kind Regards, Christian Witts -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
On Tue, Feb 23, 2010 at 1:13 PM, Giorgio anothernetfel...@gmail.com wrote: I have an update: I can easily undertand why this example doesn't work: def nochange(x): x = 0 y = 1 nochange(y) print y # Prints out 1 X is a local variable, and only gets modified in the function, that doesn't return any value. But it's very difficult for me to understand WHY this works: def change(some_list): some_list[1] = 4 x = [1,2,3] change(x) print x # Prints out [1,4,3] some_list is a local list, isn't it? Maybe i can't have lists that are only existing in a function? Here is what happens, as I understand it: When you enter the function, a new name (x, in your case) is created in the local scope. That name points to the object you supplied when you called the function (an integer object, with a value of 1). the x = 0 statement creates a new object, and has the name x now pointing to this new object. The old integer object still exists, and y still points to it. This is why the global y name is not affected by the change in x Now, in the second example, the same thing basically happens. A new name is created and pointed at the object you supplied. However, the statement some_list[1] = 4 is different from the assignment, in that it doesn't create a new object; It modifies the existing one. Since the global and local names both point to the same object, the change you make is reflected in both. You can of course create a new list object, so that the original is not affected: def nochange(some_list): # create shallow copy of list. NOTE: Shallow copies may still bite you if you change the list members. some_list = some_list[:] some_list[1] = 2 x = [1, 2, 3] nochange(x) x [1, 2, 3] HTH, Hugo ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
Thankyou Hugo! Ok, so i think the key is of my problem is that when doing X = 0 i'm creating a new object, that only exist in the local namespace. BUT, when using a list as a parameter for a function i'm only giving it a new name, but the object it's referring to it's always the same, and is in the global namespace. Right? 2010/2/23 Hugo Arts hugo.yo...@gmail.com On Tue, Feb 23, 2010 at 1:13 PM, Giorgio anothernetfel...@gmail.com wrote: I have an update: I can easily undertand why this example doesn't work: def nochange(x): x = 0 y = 1 nochange(y) print y # Prints out 1 X is a local variable, and only gets modified in the function, that doesn't return any value. But it's very difficult for me to understand WHY this works: def change(some_list): some_list[1] = 4 x = [1,2,3] change(x) print x # Prints out [1,4,3] some_list is a local list, isn't it? Maybe i can't have lists that are only existing in a function? Here is what happens, as I understand it: When you enter the function, a new name (x, in your case) is created in the local scope. That name points to the object you supplied when you called the function (an integer object, with a value of 1). the x = 0 statement creates a new object, and has the name x now pointing to this new object. The old integer object still exists, and y still points to it. This is why the global y name is not affected by the change in x Now, in the second example, the same thing basically happens. A new name is created and pointed at the object you supplied. However, the statement some_list[1] = 4 is different from the assignment, in that it doesn't create a new object; It modifies the existing one. Since the global and local names both point to the same object, the change you make is reflected in both. You can of course create a new list object, so that the original is not affected: def nochange(some_list): # create shallow copy of list. NOTE: Shallow copies may still bite you if you change the list members. some_list = some_list[:] some_list[1] = 2 x = [1, 2, 3] nochange(x) x [1, 2, 3] HTH, Hugo -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
On Tue, Feb 23, 2010 at 2:28 PM, Giorgio anothernetfel...@gmail.com wrote: Thankyou Hugo! Ok, so i think the key is of my problem is that when doing X = 0 i'm creating a new object, that only exist in the local namespace. BUT, when using a list as a parameter for a function i'm only giving it a new name, but the object it's referring to it's always the same, and is in the global namespace. Right? Well, mostly, yes. It's important to see that it's not so much the objects that live in namespaces, it's the names (otherwise they would be called object-spaces, yes?). The objects do not live inside a namespace, but are in a conceptually separate place altogether. A name lives in a namespace, and can only be referenced inside that space. An object can be referenced from anywhere, as long as you have a name that points to it. So, when you're doing x = 0, you're creating a new object, and the name x (in the local namespace) points to that object. That doesn't mean the object itself is confined to the local namespace. You could write 'return x', which allows you to have a name in the global namespace point to that same object. Hugo ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
Ok Hugo, so, going back to your example: def nochange(some_list): # create shallow copy of list. NOTE: Shallow copies may still bite you if you change the list members. some_list = some_list[:] some_list[1] = 2 Here we've created a new list. It's an object in the global object-space :) but i can't access it outside the function because i don't have a name referring to it in the global namespace. Right? And, please let me ask a question: Kent told that nested_namespace(s) are default in python 2.6. And i found a line confirming this in py2.6 library. But, what about python 2.5 that as you know is the default on linux? Thankyou Giorgio 2010/2/23 Hugo Arts hugo.yo...@gmail.com On Tue, Feb 23, 2010 at 2:28 PM, Giorgio anothernetfel...@gmail.com wrote: Thankyou Hugo! Ok, so i think the key is of my problem is that when doing X = 0 i'm creating a new object, that only exist in the local namespace. BUT, when using a list as a parameter for a function i'm only giving it a new name, but the object it's referring to it's always the same, and is in the global namespace. Right? Well, mostly, yes. It's important to see that it's not so much the objects that live in namespaces, it's the names (otherwise they would be called object-spaces, yes?). The objects do not live inside a namespace, but are in a conceptually separate place altogether. A name lives in a namespace, and can only be referenced inside that space. An object can be referenced from anywhere, as long as you have a name that points to it. So, when you're doing x = 0, you're creating a new object, and the name x (in the local namespace) points to that object. That doesn't mean the object itself is confined to the local namespace. You could write 'return x', which allows you to have a name in the global namespace point to that same object. Hugo -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
On Tue, Feb 23, 2010 at 1:55 PM, Giorgio And, please let me ask a question: Kent told that nested_namespace(s) are default in python 2.6. And i found a line confirming this in py2.6 library. But, what about python 2.5 that as you know is the default on linux? Yes, since 2.2 nested namespaces have been standard. Kent ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
Ok, thankyou. So, in other words, i must pay attention to what i set as default value in a function. I should NEVER set empty lists as default values. The guide i've linked says To learn more about this, you should read the documentation and look for the difference between *identity* and *equality*. . Can you please help me also with that? Where can i read what is the difference? And, i have some difficulties understanding the other strange example in that howto. Just scroll down to: However, the *point* is that the value of x is picked up from the *environment* at the time when the function is defined. How is this useful? Let’s take an example — a function which composes two other functions. He is working on a function that compose other 2 functions. This is the final solution def compose(fun1, fun2): def inner(x, fun1=fun1, fun2=fun2): return fun1(fun2(x)) return inner But also tries to explain why this example: # Wrong version def compose(fun1, fun2): def inner(x): return fun1(fun2(x)) return inner def fun1(x): return x + world! def fun2(x): return Hello, sincos = compose(sin,cos) # Using the wrong version x = sincos(3) Won't work. Now, the problem is that the inner function gets fun1 and fun2 from other 2 functions. My question is: why? inner is a sub-function of compose, where fun1 and fun2 are defined. Giorgio 2010/2/21 Steven D'Aprano st...@pearwood.info On Mon, 22 Feb 2010 03:00:32 am Giorgio wrote: Hi, do you know if there is a way so that i can get multiple values from a function? For example: def count(a,b): c = a + b d = a - b How can I return the value of C and D? Return a tuple of c and d: def count(a, b): ... c = a + b ... d = a - b ... return c, d ... t = count(15, 11) t (26, 4) You can also unpack the tuple immediately: x, y = count(15, 11) x 26 y 4 Then, i have another question: i've read, some time ago, this guide http://hetland.org/writing/instant-python.html, skipping the object-related part. Now i've started reading it, and have found something strange: just go where it says Of course, now you know there is a better way. And why don’t we give it the default value of [] in the first place? Because of the way Python works, this would give all the Baskets the same empty list as default contents.. Can you please help me understanding this part? When you declare a default value in a function like this: def f(a, b, c=SOMETHING): the expression SOMETHING is calculated once, when the function is defined, and *not* each time you call the function. So if I do this: x = 1 y = 2 def f(a, b, c=x+y): return a+b+c the default value for c is calculated once, and stored inside the function: f(0, 0) 3 Even if I change x or y: x = f(0, 0) 3 So if I use a list as a default value (or a dict), the default is calculated once and stored in the function. You can see it by looking at the function's defaults: def f(alist=[]): ... alist.append(1) ... return alist f.func_defaults[0] [] Now, call the function without an argument: f() [1] f() [1, 1] f() [1, 1, 1] f.func_defaults[0] [1, 1, 1] How is this happening? Because every time you call the function, it appends 1 to the argument. If you don't supply an argument, it appends 1 to the default, changing it in place. Why doesn't the same thing happen here? def g(x=0): ... x += 1 ... return x ... g.func_defaults[0] 0 g() 1 g() 1 g.func_defaults[0] 0 The answer is that ints are immutable: you can't change their value. When you do x+=1, it doesn't modify the int 0 in place, turning it into 1. It leaves 0 as zero, and gives you a new int equal to one. So the default value stored in the function never changes. The difference boils down to immutable objects, which can't be changed in place, and mutable objects, which can. Immutable: ints, floats, strings, tuples, frozensets Mutable: lists, dicts, sets, most custom classes -- Steven D'Aprano ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
On Mon, Feb 22, 2010 at 9:13 AM, Giorgio anothernetfel...@gmail.com wrote: And, i have some difficulties understanding the other strange example in that howto. Just scroll down to: However, the point is that the value of x is picked up from the environment at the time when the function is defined. How is this useful? Let’s take an example — a function which composes two other functions. He is working on a function that compose other 2 functions. This is the final solution def compose(fun1, fun2): def inner(x, fun1=fun1, fun2=fun2): return fun1(fun2(x)) return inner But also tries to explain why this example: # Wrong version def compose(fun1, fun2): def inner(x): return fun1(fun2(x)) return inner def fun1(x): return x + world! def fun2(x): return Hello, sincos = compose(sin,cos) # Using the wrong version x = sincos(3) Won't work. Now, the problem is that the inner function gets fun1 and fun2 from other 2 functions. My question is: why? inner is a sub-function of compose, where fun1 and fun2 are defined. It does work: In [6]: def compose(fun1, fun2): ...: def inner(x): ...: return fun1(fun2(x)) ...: return inner ...: In [7]: def fun1(x): ...: return x + world! ...: In [8]: def fun2(x): ...: return Hello, ...: In [9]: from math import sin, cos In [10]: sincos = compose(sin,cos) # Using the wrong version In [11]: In [12]: x = sincos(3) In [13]: In [14]: x Out[14]: -0.8360218615377305 That is a very old example, from python 2.1 or before where nested scopes were not supported. See the note A Note About Python 2.1 and Nested Scopes - that is now the default behaviour. Kent ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
Ahah Kent this is amazing. I was reading the ITALIAN http://www.python.it/doc/articoli/instpy-0.html version of that guide that is not updated. But, when i decided to post there i've posted the link of the guide in english, but actually that's not what i've readen. Ok, so in the new python version it works. Tha manual also states that: The features recognized by Python 2.6 are unicode_literals, print_function, absolute_import, division, generators, nested_scopes andwith_statement. generators, with_statement, nested_scopes are redundant in Python version 2.6 and above because they are always enabled. Kent, as i'm learning py, can you please spend some words on nested scopes? What are them? And why are supported anymore? And, if i'm not asking you too much: can you plase post an example on how that function can return HEllo World in py 2.6? Thankyou! Giorgio 2010/2/22 Kent Johnson ken...@tds.net On Mon, Feb 22, 2010 at 9:13 AM, Giorgio anothernetfel...@gmail.com wrote: And, i have some difficulties understanding the other strange example in that howto. Just scroll down to: However, the point is that the value of x is picked up from the environment at the time when the function is defined. How is this useful? Let’s take an example — a function which composes two other functions. He is working on a function that compose other 2 functions. This is the final solution def compose(fun1, fun2): def inner(x, fun1=fun1, fun2=fun2): return fun1(fun2(x)) return inner But also tries to explain why this example: # Wrong version def compose(fun1, fun2): def inner(x): return fun1(fun2(x)) return inner def fun1(x): return x + world! def fun2(x): return Hello, sincos = compose(sin,cos) # Using the wrong version x = sincos(3) Won't work. Now, the problem is that the inner function gets fun1 and fun2 from other 2 functions. My question is: why? inner is a sub-function of compose, where fun1 and fun2 are defined. It does work: In [6]: def compose(fun1, fun2): ...: def inner(x): ...: return fun1(fun2(x)) ...: return inner ...: In [7]: def fun1(x): ...: return x + world! ...: In [8]: def fun2(x): ...: return Hello, ...: In [9]: from math import sin, cos In [10]: sincos = compose(sin,cos) # Using the wrong version In [11]: In [12]: x = sincos(3) In [13]: In [14]: x Out[14]: -0.8360218615377305 That is a very old example, from python 2.1 or before where nested scopes were not supported. See the note A Note About Python 2.1 and Nested Scopes - that is now the default behaviour. Kent -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
[Tutor] Functions returning multiple values
Hi, do you know if there is a way so that i can get multiple values from a function? For example: def count(a,b): c = a + b d = a - b How can I return the value of C and D? Then, i have another question: i've read, some time ago, this guide http://hetland.org/writing/instant-python.html, skipping the object-related part. Now i've started reading it, and have found something strange: just go where it says Of course, now you know there is a better way. And why don’t we give it the default value of [] in the first place? Because of the way Python works, this would give all the Baskets the same empty list as default contents.. Can you please help me understanding this part? Thankyou Giorgio -- -- AnotherNetFellow Email: anothernetfel...@gmail.com ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
Re: [Tutor] Functions returning multiple values
On Mon, 22 Feb 2010 03:00:32 am Giorgio wrote: Hi, do you know if there is a way so that i can get multiple values from a function? For example: def count(a,b): c = a + b d = a - b How can I return the value of C and D? Return a tuple of c and d: def count(a, b): ... c = a + b ... d = a - b ... return c, d ... t = count(15, 11) t (26, 4) You can also unpack the tuple immediately: x, y = count(15, 11) x 26 y 4 Then, i have another question: i've read, some time ago, this guide http://hetland.org/writing/instant-python.html, skipping the object-related part. Now i've started reading it, and have found something strange: just go where it says Of course, now you know there is a better way. And why don’t we give it the default value of [] in the first place? Because of the way Python works, this would give all the Baskets the same empty list as default contents.. Can you please help me understanding this part? When you declare a default value in a function like this: def f(a, b, c=SOMETHING): the expression SOMETHING is calculated once, when the function is defined, and *not* each time you call the function. So if I do this: x = 1 y = 2 def f(a, b, c=x+y): return a+b+c the default value for c is calculated once, and stored inside the function: f(0, 0) 3 Even if I change x or y: x = f(0, 0) 3 So if I use a list as a default value (or a dict), the default is calculated once and stored in the function. You can see it by looking at the function's defaults: def f(alist=[]): ... alist.append(1) ... return alist f.func_defaults[0] [] Now, call the function without an argument: f() [1] f() [1, 1] f() [1, 1, 1] f.func_defaults[0] [1, 1, 1] How is this happening? Because every time you call the function, it appends 1 to the argument. If you don't supply an argument, it appends 1 to the default, changing it in place. Why doesn't the same thing happen here? def g(x=0): ... x += 1 ... return x ... g.func_defaults[0] 0 g() 1 g() 1 g.func_defaults[0] 0 The answer is that ints are immutable: you can't change their value. When you do x+=1, it doesn't modify the int 0 in place, turning it into 1. It leaves 0 as zero, and gives you a new int equal to one. So the default value stored in the function never changes. The difference boils down to immutable objects, which can't be changed in place, and mutable objects, which can. Immutable: ints, floats, strings, tuples, frozensets Mutable: lists, dicts, sets, most custom classes -- Steven D'Aprano ___ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
