Cool, thanks! It makes the solution much more readable.

czw., 29 sie 2019, 12:47 użytkownik Roman Perepelitsa <
roman.perepelitsa@gmail.com> napisał:

> On Thu, Aug 29, 2019 at 2:51 AM Sebastian Gniazdowski
> <sgniazdowski@gmail.com> wrote:
> >
> > myarr=( "${(@)${myarr[@]/(#m)*/$(( msfunc(${(q)MATCH})
> ))$REPLY}/(#s)0/}" )
>
> I usually use ${foo+} to expand foo for its side effects and to ignore
> the substituted value.
>
>     myarr=( "${(@)${myarr[@]/(#m)*/${$((msfunc(${(q)MATCH})))+}$REPLY}}" )
>
> This way you can discard the substituted value without knowing what it is.
>
> Roman.
>