Re: Local variables overriding global constants

[Chris F.A. Johnson]

On Wed, 3 Apr 2013, Nikolai Kondrashov wrote:

> On 04/03/2013 10:53 AM, Chris Down wrote:
> > On 2013-04-03 10:50, Nikolai Kondrashov wrote:
> > > On 04/03/2013 10:43 AM, Chris F.A. Johnson wrote:
> > > > On Wed, 3 Apr 2013, Nikolai Kondrashov wrote:
> > > > > I.e. this:
> > > > >
> > > > > bash -c 'declare -r v; a() { declare -r v; }; a'
> > > > >
> > > > > Results in:
> > > > >
> > > > > bash: line 0: declare: v: readonly variable
> > > >
> > > > It doesn't work because you are trying to redefine an existing
> > > > readonly variable.
> > >
> > > Yes, but I'm explicitly redefining it locally, only for this function.
> > > And this works for variables previously defined in the calling function.
> >
> > You're not redefining it locally, you are unsuccessfully trying to override 
> > a
> > global.
>
> How is this different?
>
>    bash -c 'declare v=1; function a() { declare v=2; }; a; echo "$v"'

   In a() you created a new, local variable, which you could do in
   this instance because it is not readonly.

> > > > > While this works:
> > > > >
> > > > > bash -c 'a() { declare -r v; }; b() { declare -r v; a; }; b'
> > > >
> > > > It works because both instances are local to a function and don't
> > > > exist outside their own functions.
> > >
> > > Not true.
> > >
> > > This:
> > >
> > >      bash -c 'a() { echo "$v"; }; b() { declare -r v=123; a; }; b'
> > >
> > > Produces this:
> > >
> > >      123
> >
> > That is *inside* the function, not *outside* the function.
>
> I'd say that "v" is declared outside function "a", where it is accessed, or 
> do
> you mean that there is no concept of separate functions in Bash and all
> "functions" are just one function?

   a() is a child of b(), so it can see its variables.

--
   Chris F.A. Johnson, <http://cfajohnson.com/>
   Author:
   Pro Bash Programming: Scripting the GNU/Linux Shell (2009, Apress)
   Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)