I wanted to create a script that executes the ls command with the provided arguments. Why isn't it working? I tried running it with for example -R, then it echoes (I echoed the same thing for comparison):

    ls "-R"

but says that it cannot access "-R": no such file or directory. But when I execute ls "-R" in the terminal, it works.

The script itself:


while [[ $# -gt 0 ]]
    text+=' "'


echo 'ls "-R"'

echo $text
  • ls "$@" will run ls with all arguments in the positional parameters.
    – mikeserv
    Mar 23, 2014 at 19:45

2 Answers 2


The wrong quoting is the problem. " does not quote (and, which is the problem here, does not get removed!) if it is contained in a variable. Exception: eval $text

You should assign the parameters to an array and use ls "${vars[@]}" or ls "$@" instead.

Compare the two calls:

set -x
ls -l
ls "-l"
set +x
  • FYI, I added "eval" before "$text" and it's working! Thanks, man! And the one below, thanks too!
    – Darge Elar
    Mar 23, 2014 at 19:22

That's not how quoting work in bash. When entering

ls "-R"

quotes removal happens on the command, so the command becomes

ls -R

If, on the other hand, if you set

text='ls "-R"'

and run $text, the quote removal does not happen. See man bash under EXPANSIONS:

After the preceding expansions, all unquoted occurrences of the characters \, ', and " that did not result from one of the above expansions are removed.

" here resulted from parameter expansion, so it is not removed.

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