why functions in modules need 'global foo' for integer foo but not dictionary foo?
At top of a module I have an integer like so... foo = 4 In a function in that module I know I need to do 'global foo' to get at the value 4. ... IIRC, for dictionaries you DO NOT have this issue? Why this scope problem with integers but not dictionaries? Chris -- http://mail.python.org/mailman/listinfo/python-list
Re: why functions in modules need 'global foo' for integer foo but not dictionary foo?
[EMAIL PROTECTED] wrote:
At top of a module I have an integer like so...
foo = 4
In a function in that module I know I need to do 'global foo' to get at
the value 4.
...
You don't. You need the global declaration to *set* the value of foo
in the module's namespace. When getting the value of foo within the
function, if it's not been defined in the function's local namespace, it
automatically looks in the global namespace with or without the global
foo declaration. When assigning a value to the name foo within the
function, however, without the global declaration, the value will only
be assigned to the name foo within the local namespace.
http://docs.python.org/ref/naming.html
IIRC, for dictionaries you DO NOT have this issue?
Why this scope problem with integers but not dictionaries?
I presume you are trying code like the following:
foo = 4
bar = {}
def fun1():
foo = 5
def fun2():
bar['baz'] = 4
Since integers are immutable, all that fun1() does is assign 5 to the
name foo within fun1()'s namespace and doesn't touch the module-level
namespace.
fun2(), on the other hand, only gets the dictionary from the
module-level namespace with the name bar. Then it modifies that
dictionary (since it's mutable). It does not try to assign a new object
to the name bar.
I hope that's clear, but I'm pretty tired right now, and it may not be.
--
Robert Kern
[EMAIL PROTECTED]
In the fields of hell where the grass grows high
Are the graves of dreams allowed to die.
-- Richard Harter
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Re: why functions in modules need 'global foo' for integer foo but not dictionary foo?
Robert Kern wrote:
[EMAIL PROTECTED] wrote:
At top of a module I have an integer like so...
foo = 4
In a function in that module I know I need to do 'global foo' to get at
the value 4.
...
I presume you are trying code like the following:
foo = 4
bar = {}
def fun1():
foo = 5
def fun2():
bar['baz'] = 4
Since integers are immutable, all that fun1() does is assign 5 to the
name foo within fun1()'s namespace and doesn't touch the module-level
namespace.
Hmm this is obscure to me also, what mutables has to do with this
problem?A binding is always mutable.
fun2(), on the other hand, only gets the dictionary from the
module-level namespace with the name bar. Then it modifies that
dictionary (since it's mutable). It does not try to assign a new object
to the name bar.
Probably the point is modifing/rebinding the bound values and not
rebind them.
Generally if you need to use globals for inter-instance/class
communication, I suggest to define a new namespace where
to put global bindings:
class Globals:
foo=4
spam={}
for lexical coherence you can put also funtions there
class Globals:
foo=4
spam={}
@staticmethod
def func():pass
then you always refer to them with Globals.foo,Globals.spam,Globals.func
or with Module.Globals from outside.
Finally consider to use singletons as a more common approach.
Ciao
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Re: why functions in modules need 'global foo' for integer foo but not dictionary foo?
[EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] At top of a module I have an integer like so... foo = 4 In a function in that module I know I need to do 'global foo' to get at the value 4. Actually, you don't need a global foo statement to _access_ the value. You do need a global foo to _rebind_ the value. ... IIRC, for dictionaries you DO NOT have this issue? Why this scope problem with integers but not dictionaries? Telling an object to mutate itself doesn't rebind the object. so, if you wanted to say: foo = 5 and have it work at the module level, you'd need a global foo statement, while foo[bar] = spam doesn't. The dictionary foo isn't rebound, all that's happening is that its state is being changed (that is, the key and value is being added to the dictionary). HTH John Roth Chris -- http://mail.python.org/mailman/listinfo/python-list
