David, thank you.
Below is a minor modification of your code, which now fills in an
array instead of using stdout:
----cut-here----
function members
{
typeset nval
typeset -r IFS=$'\n'
nameref var=$1
nameref out=$2
typeset -a out
for nval in $(print -v var) ; do
case "$nval" in
$'\t\t'*)
;;
*=*)
nval="${nval%%=*}"
out+=( "${nval##*[$' \t']}" )
;;
esac
done
return 0
}
compound x=(
a=1
typeset -a ar=( 8 9 10 )
b=2
integer b_i=2
i="bello=hello cello=music ("
b='typeset s="hello"'
b_newline=$'typeset\ns="hello"'
j=fish
compound c=(
typeset -a ar=( 1 2 3 )
d=6
e=9
)
)
members x out_ar
print -v out_ar
----cut-here----
Olga
On Tue, Mar 22, 2011 at 8:50 PM, David Korn <[email protected]> wrote:
> cc: [email protected]
> Subject: Re: Re: [ast-users] Enumerate member variable names in compound
> variable?
> --------
>
>> David, your example doesn't work for nested compound variables, i.e.
>> 'compound x=( a=1 b=2 c=3 compound d=( a=1 b=2 compund z=( y=3 )) )'.
>> I tried it myself and did not found a simple way to extract just the
>> names of the members of the top level and not any level below.
>>
>> Olga
>>
>
> The following function should output only the top level.
> =====================cut here===============================
> function members
> {
> typeset nval IFS=$'\n' blank=$' \t'
> nameref var=$1
> for nval in $(print -v var)
> do case $nval in
> $'\t\t'*) ;;
> *=*) nval=${nval%%=*}
> print "${nval##*[$blank]}";;
> esac
> done
> }
> =====================cut here===============================
>
> David Korn
> [email protected]
>
--
, _ _ ,
{ \/`o;====- Olga Kryzhanovska -====;o`\/ }
.----'-/`-/ [email protected] \-`\-'----.
`'-..-| / http://twitter.com/fleyta \ |-..-'`
/\/\ Solaris/BSD//C/C++ programmer /\/\
`--` `--`
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