[Python-ideas] nonlocal functions

[MRAB]

Bruce Leban wrote:
> On Tue, Oct 20, 2009 at 4:04 AM, Nick Coghlan <ncoghlan at gmail.com 
> <mailto:ncoghlan at gmail.com>> wrote:
> 
>     The difference lies in the fact that in C, object references are
>     non-local by default - you have to declare them explicitly in the
>     current scope to make them local.
> 
>     Accordingly, I find the idea of a new function-like construct where all
>     non-argument variable references are nonlocal by default to be a
>     potentially interesting one.
> 
> 
> Hmm. That's an interesting idea.
> 
> I don't know enough about internals. It might be possible to do this 
> with a decorator
> 
> def foo():
>     a = 1
>     @allnonlocal
>     def bar():
>         a = 2
>     bar()
>     return a
> 
> foo() => 2
> 
> However, even if possible, that would make everything nonlocal with no 
> way to pick which ones. Here's another idea:
> 
> def foo():
>     a = 1
>     b = 3
>     nonlocal def bar():
>         local b
>         a = 2
>         b = 4
>     return (a,b)
> 
> foo() => (2, 3)
> 
> Prepending nonlocal to a function definition is equivalent to applying 
> nonlocal to every variable referenced in that function unless the 
> variable is declared local. For backward compatibility, the local 
> statement would only be only recognized inside nonlocal functions. (Thus 
> if you have a function that uses 'local' as a variable it would not need 
> to be changed unless you decided to stick nonlocal in front of it.)
> 
> Alternative syntax:
>     def bar() nonlocal:
> 
I'd prefer:

def foo():
     a = 1
     b = 3
     def bar():
         nonlocal *
         local b
         a = 2
         b = 4
     return (a,b)

if a later 'local' can override an earlier 'nonlocal', or:

def foo():
     a = 1
     b = 3
     def bar():
         local b
         nonlocal *
         a = 2
         b = 4
     return (a,b)

if a 'nonlocal' can act as a catch-all for any names not previously
mentioned.