10/12/23 14:11, Omar Polo пишет:
> On 2023/10/11 19:10:43 +0000, Klemens Nanni <k...@openbsd.org> wrote:
>> On Wed, Oct 11, 2023 at 03:44:47PM +0200, Omar Polo wrote:
>>> -@@ -27,7 +27,7 @@ export LEIN_HOME="${LEIN_HOME:-"$HOME/.lein"}"
>>> +@@ -48,19 +48,21 @@ function make_native_path {
>>> + # usage : add_path PATH_VAR [PATH]...
>>> + function add_path {
>>> + local path_var="$1"
>>> ++ local val=""
>>> + shift
>>> + while [ -n "$1" ];do
>>> + #
>>> http://bashify.com/?Useful_Techniques:Indirect_Variables:Indirect_Assignment
>>> +- if [[ -z ${!path_var} ]]; then
>> The comment's link just tells me the domain is for sale and bash(1) does
>> not document ${!name} (only ${!prefix*} and other forms); not sure what
>> they do here, but it looks like a simple empty check suffices?
> It's documented here: apparentely
>
> https://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
>
> : If the first character of parameter is an exclamation point (!), and
> : parameter is not a nameref, it introduces a level of indirection
>
> The idea of this `add_path' is to take a variable name as the first
> argument and append the rest of the arguments.
Thanks for the clue.
>>> +- export ${path_var}="$(make_native_path "$1")"
>>> ++ if [ -z "$val" ]; then
>> ksh(1) has [[ and you should stick to that, there quoting is not needed
>> compared to [ -- see the manual for differences.
> I know the difference between [ and [[. I try to stick with the posix
> sh subset even in a blatantly bash/ksh evironment, but can change to
> the fancy [[ if you prefer.
It just looked odd to downgrade from [[ to [ when you
use bash/ksh in the shebang already and other lines in the
same script already use [[, that's all.
Your call.
>>> ++ val="$(make_native_path "$1")"
>>> + else
>>> +- export ${path_var}="${!path_var}${delimiter}$(make_native_path
>>> "$1")"
>>> ++ val="${val}${delimiter}$(make_native_path "$1")"
>>> + fi
>>> + shift
>>> + done
>>> ++ export ${path_var}="$val"
>> Instead of deferring the export, does it work if you just remove the two !
>> and leave the rest as-is?
> Nope, because we loose the initial value.
>
> The alternative would have been to use eval to get the value of
> $path_var (not the content of path_var but the content of the variable
> named "$path_var").
No need to pull out eval (:
Your original diff is OK kn, regardless of brackets.
