On Mon, Jun 26, 2023, 14:08 Martin D Kealey <mar...@kurahaupo.gen.nz> wrote:
> hmm, declare -p used to print an array like
> ˋˋˋ
> declare -a array='([0]="value" [1]="value")'
> ˋˋˋ
> At some stage declare -p stopped printing the extra outer quotes, so that
> my approach also stops working.
>
quotes matter to input
different data == different quotes
On Mon, 26 Jun 2023, 19:32 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
> >
> > --
> > Kerin Millar
> >
>
