Question

What I’d like to do is take, as an input to a function, a line that may include quotes (single or double) and echo that line exactly as it was provided to the function. For instance:

function doit {
   printf "%s " ${@} 
   eval "${@}"
   printf " # [%3d]\n" ${?}
}

Which, given the following input

doit VAR=42
doit echo 'single quote $VAR'
doit echo "double quote $VAR"

Yields the following:

VAR=42  # [  0]
echo single quote $VAR  # [  0]
echo double quote 42  # [  0]

So the semantics of the variable expansion are preserved as I’d expect, but I can not get the exact format of the line as it was provided to the function. What I’d like is to have doit echo 'single quote $VAR' result in echo 'single quote $VAR'.

I’m sure this has to do with bash processing the arguments before they are passed to the function; I’m just looking for a way around that (if possible).

Edit

So what I had intended was to shadow the execution of a script while providing an exact replica of the execution that could be used as a diagnostic tool including exit status of each step.

While I can get the desired behavior described above by doing something like

while read line ; do 
   doit ${line}
done < ${INPUT}

That approach fails in the face of control structures (i.e. if, while, etc). I thought about using set -x but that has it’s limitations as well: " becomes ' and exit status is not visible for commands that fail.

Answer 1

The reason this happens is because bash interprets the arguments, as you thought. The quotes simply aren’t there any more when it calls the function, so this isn’t possible. It worked in DOS because programs could interpret the command line themselves, not that it helps you!