Re: Enable compgen even when programmable completions are not available?

[alex xmb ratchev]

On Mon, Jun 26, 2023, 11:33 Kerin Millar <kfm@plushkava.net> wrote:

> On Mon, 26 Jun 2023 17:09:47 +1000
> Martin D Kealey <martin@kurahaupo.gen.nz> wrote:
>
> > Hi Eli
> >
> > How about using the shell itself to parse the output of "typeset" (an
> alias
> > for "declare"), but redefining "declare" to do something different. This
> is
> > a bit verbose but it works cleanly:
> >
> > ```
> > (
> >   function declare {
> >     while [[ $1 = -* ]] ; do shift ; done
> >     printf %s\\n "${@%%=*}"
> >   }
> >   eval "$( typeset -p )"
> > )
> > ```
>
> Unfortunately, this is defective.
>
> $ bash -c 'declare() { shift; printf %s\\n "${1%%=*}"; }; eval "declare -a
> BASH_ARGC=()"'; echo $?
> 1
>
> In fact, bash cannot successfully execute the output of declare -p in full.
>
> $ declare -p | grep BASH_ARGC
> declare -a BASH_ARGC=([0]="0")
> $ declare -a BASH_ARGC=([0]="0"); echo $? # echo is never reached
>
> While it is understandable that an attempt to assign to certain shell
> variables would be treated as an error, the combination of not printing a
> diganostic message and inducing a non-interactive shell to exit is rather
> confusing. Further, declare is granted special treatment, even after having
> been defined as a function (which might be a bug).
>
> $ bash -c 'declare() { shift; printf %s\\n "${1%%=*}"; }; eval "declare -a
> BASH_ARGC=()"'; echo $?
> 1
>
> $ bash -c 'declare() { shift; printf %s\\n "${1%%=*}"; }; eval "declare -a
> BASH_ARG=()"'; echo $?
> BASH_ARG
> 0
>
> $ bash -c 'f() { shift; printf %s\\n "${1%%=*}"; }; eval "f -a
> BASH_ARGC=()"'; echo $?
> bash: eval: line 1: syntax error near unexpected token `('
> bash: eval: line 1: `f -a BASH_ARGC=()'
> 2
>
> $ bash -c 'f() { shift; printf %s\\n "${1%%=*}"; }; eval "f -a
> BASH_ARG=()"'; echo $?
> bash: eval: line 1: syntax error near unexpected token `('
> bash: eval: line 1: `f -a BASH_ARG=()'
> 2
>

you forgot
see u cmd foo bar=()
u still need as always escape ( and )

bash-5.2$ bash -c $'f() { shift; printf %s\\n "${1%%=*}"; }; eval "f -a
BASH_ARG=\'()\'"'; echo $?
BASH_ARGn0
bash-5.2$

-- 
> Kerin Millar
>
>