eval bash function arguments without misescaping - bash

Question

In a Bash script, I want to create a function that wraps commands, printing them before executing them.
So, in a script command like this:
mkdir -p "~/new/dir/tree/"
rsync -e 'ssh -p 22' -av "src/" "user#${HOST}:dest/"
I could put a command in front of the others like this:
run mkdir -p "~/new/dir/tree/"
run rsync -e 'ssh -p 22' -av "src/" "user#${HOST}:dest/"
I define the function run as:
run (){
echo -e "FANCY FORMATING CMD> $# FANCY FORMAT ENDING"
eval "$#"
return $?
}
It works fine for most cases. When I need the command in front of run it should work with it or without it whether you remove the run part or unset the run. I mean, all three commands should work exactly the same way:
run original_comand arg1 "arg2 subarg2" etc;
unset run;
run original_comand arg1 "arg2 subarg2" etc;
original_comand arg1 "arg2 subarg2" etc;
But if I try with this line:
run rsync -e 'ssh -p 22' -av "src/" "user#${HOST}:dest/"
the -e argument became unquoted and the command actually run is
run rsync -e ssh -p 22 -av src/ user#${HOST}:dest/
If I escape the quote on the -e argument, it may work with the run command, but it would not work without it.
As far as I could read from Bash documentation, the "$#" should work, but clearly I'm missing something here.

Answer 1

The arguments you pass to run already underwent all sorts of expansion when they were passed to run, and are also already properly positioned, so you don't need eval to invoke the command in $1. In fact, it is an error to use eval in this case (which erroneously applies expansions and word-splitting of sorts yet another time) as you can tell by the plenty error messages you get.
The proper way to invoke the command in $1 and passing it all the arguments in "$#" (except for the first, i.e. "${#:2}"), is to simply put "$#" on a single line without eval.

Answer 2

Your arguments are not "unquoted" by calling your function. They are "unquoted" by putting them in a string and echoing it. The quotes are not part of the arguments, but special characters in the shell that prevents word-splitting (and in the case of single quotes, variable expansion).
The shell does word-splitting (which quotes protect against), then quote removal. This means that the quotes would be removed, yes, but that the arguments would not be split on quoted whitespace.
In
utility a "b c"
the utility would not get "b c" as its second argument, but b c (not b and c, but b<space>c). The same goes for your function.
This shows that you still get properly separated arguments in the function:
#!/bin/sh
run () {
printf 'Arg: %s\n' "$#"
}
HOST=example.com
run mkdir -p "~/new/dir/tree/"
echo '---'
run rsync -e 'ssh -p 22' -av "src/" "user#${HOST}:dest/"
This generates
Arg: mkdir
Arg: -p
Arg: ~/new/dir/tree/
---
Arg: rsync
Arg: -e
Arg: ssh -p 22
Arg: -av
Arg: src/
Arg: user#example.com:dest/
As you can see, ssh -p 22 is still delivered to printf as a separate argument, not three arguments.
Doing eval "$#" in your function will do the right thing. Escaping the quotes would definitely not do the right thing as it would include the quotes in the actual argument (mkdir \"dir\" would create a directory called "dir", including the quotes, and rsync -e \"ssh -p 22\" would be delivered to rsync as -e, "ssh, -p, 22").
Also, note that ~ is not expanded since it was in double quotes. Use $HOME instead.