static variables in Python?

[kj]

In <w8CdnQGPtuvdGRLVnZ2dnUVZ_r7inZ2d at comcast.com> Larry Bates <larry.bates at websafe.com`> writes:

>kj wrote:
>> Yet another noob question...
>> 
>> Is there a way to mimic C's static variables in Python?  Or something
>> like it?  The idea is to equip a given function with a set of
>> constants that belong only to it, so as not to clutter the global
>> namespace with variables that are not needed elsewhere.
>> 
>> For example, in Perl one can define a function foo like this 
>> 
>> *foo = do {
>>   my $x = expensive_call();
>>   sub {
>>     return do_stuff_with( $x, @_ );
>>   }
>> };
>> 
>> In this case, foo is defined by assigning to it a closure that has
>> an associated variable, $x, in its scope.
>> 
>> Is there an equivalent in Python?
>> 
>> Thanks!
>> 
>> kynn


>First names in Python are just that, names that point to objects.  Those objects 
>can contain any type of information including other objects.  They are NOT 
>buckets where things are stored.

>1) Names (variables in Perl/C) defined within a Python function are placed in 
>its local namespace.  They are not visible in the global namespace.

>2) Yes you can have a local name point to a global.  This is often used in 
>classes with attributes because looking up local is somewhat quicker than 
>looking up the class attribute.

>def foo():
>   x = expensive_call
>   return do_stuff_with(x())

Maybe I'm missing your point, the goal is to have a "runtime
constant" associated with the function.  In the your definition of
foo, expensive_call gets called every time that foo gets called;
this is what I'm trying to avoid!

Maybe it's easier to see what I mean with JavaScript:

function foo() {
  if (foo.x === undefined) foo.x = expensive_call();
  return do_stuff_with(foo.x);
}

Here, expensive_call is called only once (assuming it never returns
undefined).

OK, I guess that in Python the only way to do what I want to do is
with objects...

kynn
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