bash wildcard expansion even if quoted

Question

The current directory contains file1 and file2. How do you explain this output:

[root@server test]# command='*';
[root@server test]# echo $command
file1 file2

Why is * expanded? I expect it not to be because it's quoted.

Because `$command` is unquoted. Try `"$command"`, and you'll get `*`. — Benjamin W., Nov 15 '18 at 15:08
Right... but I don't really understand why it has to be quoted... — jaudo, Nov 15 '18 at 15:10
Actually I think Stephane's question is a more appropriate duplicate: https://unix.stackexchange.com/q/171346/237982 — jesse_b, Nov 15 '18 at 15:12
When the command is interpreted, it undergoes a few [expansions](https://www.gnu.org/software/bash/manual/bash.html#Shell-Expansions). Relevant in this case: parameter expansion (turns `$command` into `*`), and then filename expansion (turns `*` into filenames in the current directory). Quoting suppresses that second expansion. — Benjamin W., Nov 15 '18 at 15:13

score 1 · Accepted Answer · answered Nov 15 '18 at 15:23

command='*' does indeed as you expect set the value of $command to a literal *. However, when you run echo $command, the value of $command is expanded to *, making your command echo *. The wildcard is then expanded as per usual. To prevent this, quote the parameter you are giving to echo:

$ v='*'
$ echo "$v"
*

bash wildcard expansion even if quoted

1 Answers1