Asterisk ("*") works as ls command

Question

Can you explain why

CHAR=$(printf "\x$(printf %x 42)"); echo $CHAR

works as ls command in the current directory?

I'm trying to print a character by its decimal ASCII code. "man" what and where do I need to read? Or is there some better way to get the character in decimal code?

I'm trying to get "*" in the output.

See also [When to wrap quotes around a shell variable?](https://stackoverflow.com/questions/10067266/when-to-wrap-quotes-around-a-shell-variable) — tripleee, Apr 08 '23 at 08:19
See [ASCII Table](https://www.asciitable.com) and all should become clear.... — David C. Rankin, Apr 09 '23 at 06:22

score 1 · Answer 1 · answered Apr 08 '23 at 07:38

1

Unquoted variable expansion undergoes word splitting and filename expansion. Quote it.

CHAR='*'
echo "$CHAR"

why works as ls command in the current directory?

Because CHAR contains a star, unquoted $CHAR triggers filename expansion, which replaces it by paths in the current directory. Because it was designed that way.

answered Apr 08 '23 at 07:38

KamilCuk

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It's understandable. I apologise for not revealing the task further `shuf -n 12 -zre $CHAR` performs an asterisk. I've already escaped this in all sorts of ways. At this `shuf -n 12 -zre * shuf -n 12 -zre "*"` works without any problems. – Sadok Apr 09 '23 at 08:46

Asterisk ("*") works as ls command

1 Answers1