Re: declare -f does not output esac pattern correctly

[Emanuele Torre]

On Tue, Feb 27, 2024 at 12:23:41PM +0100, Emanuele Torre wrote:
> On Tue, Feb 27, 2024 at 04:10:06PM +0700, Robert Elz wrote:
> >     Date:        Tue, 27 Feb 2024 00:50:46 +0100
> >     From:        Emanuele Torre <torreemanuele6@gmail.com>
> >     Message-ID:  <Zd0j1s452p-Vod6H@t420>
> > 
> >   | To use esac as a pattern you need to use the (esac) syntax,
> > 
> > Or quote it
> > 
> >     'esac')
> > 
> > (or similar).
> > 
> > kre
> > 
> 
> No, then the pattern is  'esac'  not  esac.
> 
> In fact, declare -f will print 'esac' and not esac in that case; as it
> would print 'es'ac instead of easc if you had used 'es'ac.
> 
> After expansion, those patterns are both always esac, but that is
> entirely besides the point.
> 
> o/
>  emanuele6

Well, technically, even after expansion, 'esac' will be \e\s\a\c, not
esac; and 'es'ac will be \e\sac.
Anyway they will both semantically only match exactly  esac, yes.

o/
 emanuele6