469,575 Members | 1,513 Online
Bytes | Developer Community
New Post

Home Posts Topics Members FAQ

Post your question to a community of 469,575 developers. It's quick & easy.

Confused by Python and nested scoping (2.4.3)

Hi. I'm new to Python, and downloaded a Windows copy a little while
ago. I was doing some experiments with nested functions, and ran into
something strange.

This code:

def outer():
val = 10
def inner():
print val
inner()

outer()

...prints out the value '10', which is what I was expecting.

But this code..

def outer():
val = 10
def inner():
print val
val = 20
inner()
print val

outer()

...I expected to print '10', then '20', but instead got an error:

print val
UnboundLocalError: local variable 'val' referenced before assignment.

I'm thinking this is some bug where the interpreter is getting ahead of
itself, spotting the 'val = 20' line and warning me about something that
doesn't need warning. Or am I doing something wrong?

Thanks,
-Sean Givan
Apr 19 '06 #1
8 1317
There are only two scopes in Python -- global scope and function scope.

On 4/19/06, Sean Givan <ki*****@nbnet.nb.ca> wrote:
Hi. I'm new to Python, and downloaded a Windows copy a little while
ago. I was doing some experiments with nested functions, and ran into
something strange.

This code:

def outer():
val = 10
def inner():
print val
inner()

outer()

..prints out the value '10', which is what I was expecting.

But this code..

def outer():
val = 10
def inner():
print val
val = 20
inner()
print val

outer()

..I expected to print '10', then '20', but instead got an error:

print val
UnboundLocalError: local variable 'val' referenced before assignment.

I'm thinking this is some bug where the interpreter is getting ahead of
itself, spotting the 'val = 20' line and warning me about something that
doesn't need warning. Or am I doing something wrong?

Thanks,
-Sean Givan
--
http://mail.python.org/mailman/listinfo/python-list

--
Kelvie
Apr 19 '06 #2
Sean Givan wrote:
def outer():
val = 10
def inner():
print val
val = 20
inner()
print val

outer()

..I expected to print '10', then '20', but instead got an error:

print val
UnboundLocalError: local variable 'val' referenced before assignment.

I'm thinking this is some bug where the interpreter is getting ahead of
itself, spotting the 'val = 20' line and warning me about something that
doesn't need warning. Or am I doing something wrong?

Short answer: No, it's not a Python bug. If inner() must modify
variables defined in outer()'s scope, you'll need to use a containing
object. E.g.:

class Storage(object):
pass
def outer():
data = Storage()
data.val = 10
def inner():
print data.val
data.val = 20
inner()
print data.val

Long answer:

The interpreter (actually, the bytecode compiler) is indeed looking
ahead. This is by design, and is why the "global" keyword exists. See
http://www.python.org/doc/faq/progra...bles-in-python

Things get more complex than that when nested function scopes are
involved. But again, the behavior you observed is a design decision,
not a bug. By BDFL declaration, there is no "parentscope" keyword
analogous to "global". See PEP 227, specifically the "Rebinding names
in enclosing scopes" section: http://www.python.org/dev/peps/pep-0227/

Hope that helps,
--Ben

Apr 20 '06 #3
Sean Givan schrieb:
Hi. I'm new to Python
welcome
ago. I was doing some experiments with nested functions, and ran into
something strange.

This code:

def outer():
val = 10
def inner():
print val
inner()

outer()

...prints out the value '10', which is what I was expecting.

But this code..

def outer():
val = 10
def inner():
print val
val = 20
inner()
print val

outer()

...I expected to print '10', then '20', but instead got an error:

print val
UnboundLocalError: local variable 'val' referenced before assignment.

I'm thinking this is some bug where the interpreter is getting ahead of
itself, spotting the 'val = 20' line and warning me about something that
just a little carefull thought
if something that basic should really be a bug
how many thousand people would discover it daily?
doesn't need warning. Or am I doing something wrong?


yes, you can't modify it
you can do it for global namespace or local
but not inbetween

val = 0
def outer():
val = 10
def inner():
global val
val = 30
inner()
print val
outer()
10 # outer val is not changed
print val # global is modified
30

hth, Daniel

Apr 20 '06 #4

"Sean Givan" <ki*****@nbnet.nb.ca> wrote in message
news:AG*********************@ursa-nb00s0.nbnet.nb.ca...
Hi. I'm new to Python, and downloaded a Windows copy a little while
ago. I was doing some experiments with nested functions, and ran into
something strange. Experiments are good. Strange can be instructive.
.... I'm thinking this is some bug Blaming the interpreter is not so good, but amazingly common among
newcomers ;-)
where the interpreter is getting ahead of itself, .... Or am I doing something wrong?


In a sense, you got ahead of yourself. And the issue has nothing to do
with nested scopes per se. When things seem strange, try a simpler
experiment.
x=1
def f(): print x
x = 2
f()

Traceback (most recent call last):
File "<pyshell#5>", line 1, in -toplevel-
f()
File "<pyshell#4>", line 2, in f
print x
UnboundLocalError: local variable 'x' referenced before assignment

The compiler compiles functions in two passes: the first to classify names
as local or global (or nested if relevant, but not really so here), the
second to generate bytecodes which depend on that classification.

Terry Jan Reedy

Apr 20 '06 #5
Sean Givan wrote:
Hi. I'm new to Python, and downloaded a Windows copy a little while
ago. I was doing some experiments with nested functions, and ran into
something strange.

This code:

def outer():
val = 10
def inner():
print val
inner()

outer()

..prints out the value '10', which is what I was expecting.

But this code..

def outer():
val = 10
def inner():
print val
val = 20
inner()
print val

outer()

..I expected to print '10', then '20', but instead got an error:

print val
UnboundLocalError: local variable 'val' referenced before assignment.

I'm thinking this is some bug where the interpreter is getting ahead of
itself, spotting the 'val = 20' line and warning me about something that
doesn't need warning. Or am I doing something wrong?


reading the reference documentation may help:

http://docs.python.org/ref/naming.html

"If a name binding operation occurs anywhere within a code block,
all uses of the name within the block are treated as references to
the current block."

</F>

Apr 20 '06 #6
I have added some spaces guessing how the original was formatted.
See the simplified example and the explanation below...

"Sean Givan" wrote...
Hi. I'm new to Python [...] something strange.
This code:

def outer():
val = 10
def inner():
print val
inner()
outer()

..prints out the value '10', which is what I was expecting.

But this code..
def outer():
val = 10
def inner():
print val
val = 20
inner()
print val
outer()

..I expected to print '10', then '20', but instead got an error:

print val
UnboundLocalError: local variable 'val' referenced before assignment.

I'm thinking this is some bug where the interpreter is getting ahead of
itself, spotting the 'val = 20' line and warning me about something that
doesn't need warning. Or am I doing something wrong?


The simplified example of both cases can be
script a.py
---------------------------------------------
val = 10

def myFunction():
print val

myFunction()
---------------------------------------------

In this case the val is not defined inside myFunction();
therefore, it is searched in the "upper level", above
the function body. Such variable is called free variable.

script b.py
---------------------------------------------
val = 10

def myFunction():
print val
val = 20

myFunction()

---------------------------------------------

In this case the val is assigned inside the myFunction()
and it is not marked to be global. In this case Python
decides that it will be the local variable (cannot be
free variable anymore). Python insists on fact that
in one block the variable can be or free or locally
bound, but not both. This is decided during the
compilation of the module and it does not depend
on whether val = 20 assignment precedes the print val
command or not. It is decided that it will be local
inside myFunction and then the print wants to use
the variable that was not assingned yet.

pepr

P.S. I have just noticed that Terry Jan Reedy answered
similarly. Never mind... Repeat, repeat, repeat.... until
you know ;)
Apr 20 '06 #7
Kelvie Wong wrote:
There are only two scopes in Python -- global scope and function scope.


No, Python has local, nested, global and built-in scope.

Kent
Apr 20 '06 #8
>> P.S. I have just noticed that Terry Jan Reedy answered
similarly. Never mind... Repeat, repeat, repeat.... until
you know ;)


Yes, and some of us appreciate the extra examples.

rick

Apr 20 '06 #9

This discussion thread is closed

Replies have been disabled for this discussion.

Similar topics

1 post views Thread by Rob Hunter | last post: by
7 posts views Thread by Max | last post: by
134 posts views Thread by Joseph Garvin | last post: by
2 posts views Thread by Robert M. Gary | last post: by
22 posts views Thread by Kurien Mathew | last post: by
reply views Thread by suresh191 | last post: by
By using this site, you agree to our Privacy Policy and Terms of Use.