On Mon, Jun 26, 2023, 12:01 Kerin Millar <k...@plushkava.net> wrote: > On Mon, 26 Jun 2023 11:51:58 +0200 > 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 ) > > I didn't forget anything. Martin's proposal was intended to work by > evaluating the unmodified output of typeset -p. That ( and ) normally need > to be escaped simply demonstrates further that it is untenable as a > solution.
1. making bash eat func foo=() never worked without escaping ( ) .. same err msg 2. cab u paste me the specific msg / url to that declare -p .. im parser pro .. > -- > Kerin Millar >