Re: Local variables overriding global constants

[Nikolai Kondrashov]

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"'

> > > > 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?

Sincerely,
Nick