Re: Local variables overriding global constants

[Chris F.A. Johnson]

On Wed, 3 Apr 2013, Pierre Gaston wrote:
> On Wed, Apr 3, 2013 at 11:33 AM, Chris F.A. Johnson <address@hidden>wrote:
> > On Wed, 3 Apr 2013, Pierre Gaston wrote:
> >  On Wed, Apr 3, 2013 at 11:03 AM, Chris Down <address@hidden> wrote:
> > >  On 2013-04-03 11:00, Nikolai Kondrashov wrote:
> > > > >  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.
> > > >  Still Nikolai has a point.

> > > It's not clear why readonly variable can be overridden when the
> > > variable is declared readonly in the scope of an englobing
> > > function but not if it is declared readonly in the global scope.
> >
> >    If it's declared readonly in a function, the variable doesn't exist
> >    outside of that function, so it's not readonly there.
>
> I think you missed the point that "a" is called inside "b".
> See the example below
>
> >  $ bash -c 'a() {  v=2;echo "$v"; }; b () { declare -r v=1; a; echo "$v";
> > > };
> > > b'
> > > bash: v: readonly variable
>
> if v doesn't exist in "a" why does it complain that it's readonly?

   It *does* exist inside a() if a() is a child of b().

--
   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)