How do I remove newline from the last argument in bash?

Question

I'm working on wsl bash and I have the following function defined in my ~/.bashrc file:

function clip { /mnt/c/Windows/System32/clip.exe "${@%\n}"; }

However this does not remove the trailing \n from the arugument. How can I improve this to meet my specific needs?

Many thanks

EDIT: With the help of Kamil's answer, I was able to figure out what was actually going on. turns out that the spawned subprocess (cmd.exe) is adding the carriage return which is being inserted into clip.exe (because I was using pipes. ex. pwd | clip).

I was able to get the desired result with the following:

function clip { /mnt/c/Windows/System32/WindowsPowershell/v1.0/powershell.exe Set-Clipboard "${@:-$(cat)}"; }

It's a bit slow as it relies on

• powershell
• cat (external binary which should be replaced with read builtin)

But not a problem for my needs :)

Answer 1

"${@%\n}" removes a trailing n. "${@%\\n}" would remove trailing \n (literal string, not a newline).

To remove a newline character you need to pass a newline character. In Bash this syntax should work:

"${@%$'\n'}"

where $'\n' is responsible for generating an actual newline character. A literal newline will work as well:

"${@%
}"

Notes:

• The solution removes trailing newline from each argument, not only from the last (because this is how $@ works).
• In Linux its unusual to have a trailing newline in a command line argument. I don't know WSL, I don't know your usage case; I hope you know what you want^*.

---

Example:

$ set a $'b\n' c
$ printf '<%s>\n' "$@"
<a>
<b
>
<c>
$ printf '<%s>\n' "${@%$'\n'}"
<a>
<b>
<c>
$ printf '<%s>\n' "${@%
}"
<a>
<b>
<c>

---

^{* If "removing the trailing \n from the argument" is not what you want, then maybe you call the function in a (sourced?) script and the script suffers from DOS line ending(s). Check this question.}