On Mon, Jun 26, 2023, 11:51 alex xmb ratchev <fxmb...@gmail.com> wrote:
>
>
> On Mon, Jun 26, 2023, 11:33 Kerin Millar <k...@plushkava.net> wrote:
>
>> On Mon, 26 Jun 2023 17:09:47 +1000
>> Martin D Kealey <mar...@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 $?
> BASH_ARGn0
> bash-5.2$
>
by the way , what does this nonsense code supposed to do
separate between vars for future list / deletement ?
--
>> Kerin Millar
>>
>>
