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.
> >>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.
Chris
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