Je Sat, Jan 25, 2025 at 07:16:14AM +0000, Ross skribis:
>
> Configuration Information [Automatically generated, do not change]:
> Machine: x86_64
> OS: linux-gnu
> Compiler: gcc
> Compilation CFLAGS: -g -O2 -fno-omit-frame-pointer -mno-omit-leaf-
> frame-pointer -flto=auto -ffat-lto-objects -fstack-protector-strong -
> fstack-clash-protection -Wformat -Werror=format-security -fcf-
> protection -Wall
> uname output: Linux X220-Bravo 6.8.0-51-generic #52-Ubuntu SMP
> PREEMPT_DYNAMIC Thu Dec 5 13:09:44 UTC 2024 x86_64 x86_64 x86_64
> GNU/Linux
> Machine Type: x86_64-pc-linux-gnu
>
> Bash Version: 5.2
> Patch Level: 21
> Release Status: release
>
> Description:
> >From the command line
> printf "%*s\n" 80 " " | tr " " "*"
> does just what I'd expect generating a banner line of asterisks.
>
> However:
> foo=$(printf "%*s\n" 80 " " | tr " " "*")
> echo $foo
> acts very differently; it seems to perform ls or something similar.
>
> The same occurs with the more standard printf "%80s" form.
>
> Repeat-By:
> As above.
>
> Fix:
> Unknown.
>
This is no bug in bash but in your shell code. There is almost never a
reason *not* to double quote a variable substitution. Doing so will
prevent the shell from splitting the value of the variable on the
characters in $IFS (space, tab, and newline, by default), but also from
using the split-up value as filename globbing patterns, which is what
happens in your case.
Just double quote the variable expansion:
echo "$foo"
--
Andreas (Kusalananda) Kähäri
Uppsala, Sweden
.
