Re: Listing the files of a package output

[Konrad Hinsen]

Hi Guillaume,

> As the directory of the "out" output will have a shorter name, you could
> sort the names by size and take the first one:
>
> --8<---------------cut here---------------start------------->8---
> for name in $(guix build coreutils); do printf "%d\t%s\n" "${#name}" 
> "${name}"; done | sort -n | head -n 1 | cut -f 2
> --8<---------------cut here---------------end--------------->8---

Indeed, thanks! A but lengthy by my taste, but it does the job!

Cheers,
  Konrad.