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Generally, I am wondering how I can pass the arguments from a terminal command to stdout. More specifically, I am using tar as part of my backup process and I would like the parameters (especially excluded dirs) to be passed to a log file. For the excluded dirs, I tried tar --show-omitted-dirs, but that did not work as I thought.

I am new to working with bash scripts, but I read up about "$@" and figured I could write my own function

myLS() { echo "$@"; ls "$@"; }

to first echo the commands to stdout and then pass them to ls. Piping echo

myLS() { echo "$@" | tr " " "\n"; ls "$@"; }

or using printf

myLS() { printf '%s\n' "$@"; ls "$@"; }

separates these arguments on new lines, which is what I would want on top of the log file.

While this works, I am curious if there is a preferred way to go about it, which does not require me to rewrite my own function every time I want to keep a log of the arguments I used when running a command. Both general solution and solutions specific to tar are helpful.

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There is in fact no preferred way to go about doing this. The best way that I can think of, and the method I do use is exactly what you did. Create a function that first prints the paramaters, and then runs the command with those paramaters. The only improvement I can think to make is instead of using a different function for each command, you could write one function such as:

#!/bin/bash
function show_args {
    echo "$@" | tr " " "\n" >> logfile
    $@
}

This function can be called like:

show_args ls -ah ~/documents/

It will echo the arguments stacked on top of eachother to "logfile" and then run the command.

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