Store literal wildcard character in bash string

Question

Say I have a function

print_args () {
    for i in $@; do
        echo "$i"
    done
}

When I do

foo='\*'
print_args $foo

I get

\*

(with the backslash) as output.

If I change the definition of foo to foo='*' instead, I get all the files in the current directory when running print_args $foo.

So I either get the backslash included, or the * interpreted, but I don't see how to get the * literally.

The output is the same whether I include double quotes around $foo or not.

@Etan Sorry, I edited just before you finished your comment. It behaves the same double quotes or not. — Chris Middleton, Jul 22 '15 at 04:52
Oh, you also need quotes around the expansion of `$@`. `for i in "$@"`. General rule: "Always quote your variable expansions." — Etan Reisner, Jul 22 '15 at 04:54
Ah, that's it. I thought `$@` was special in that regard but I guess I was misremembering. — Chris Middleton, Jul 22 '15 at 04:58

score 1 · Accepted Answer · edited Dec 10 '15 at 23:04

1

The general rule is to quote all variables. It prevents shell expansion and splitting on spaces. So your function should look like this (quoting $@, as well as ${array[@]} splits by arguments):

print_args () {
    for i in "$@"; do
        echo "$i"
    done
}

And call it like this:

print_args "$foo"

edited Dec 10 '15 at 23:04

Josh Crozier

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answered Jul 22 '15 at 08:03

petersohn

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Store literal wildcard character in bash string

1 Answers1