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

[Kerin Millar]

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

-- 
Kerin Millar