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.
>
> How is this different?
>
> bash -c 'declare v=1; function a() { declare v=2; }; a; echo "$v"'
From `help declare':
When used in a function, `declare' makes NAMEs local, as with the `local'
command. The `-g' option suppresses this behavior.
> >>This:
> >>
> >> bash -c 'a() { echo "$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?
$v is not populated at initial runtime, it is populated when it is accessed.
Chris
pgpJ99rQ_Absd.pgp
Description: PGP signature
