How to only do quote removal in a shell?

Question

Given a string from an untrusted source, e.g.

MALICIOUS_INPUT="$(awk -F = '/^VERSION=/ {print $2}' /path/to/compromised/etc/os-release | head -n 1)"

is it possible to just apply pure shell quote removal (see Shell Command Language (IEEE Std 1003.1-2017) and Bash manual) to that string i.e. without doing variable expansions, arithmetic expansions, command substitution and similar?

This is needed, for example to parse strings from os-release files without source-ing the files.

Input	Expected result
'\"'	\"
"\""	"
'$foo${foo}$(pwd)$((1+2))'	$foo${foo}$(pwd)$((1+2))
"$foo${foo}$(pwd)$((1+2))"	$foo${foo}$(pwd)$((1+2))

Answer 1

Comparing applicability of the preexisting answers on Reading quoted/escaped arguments correctly from a string to this question:

parse_with_xargs() {
  xargs printf '%s\0' <<<"$*"
}

parse_with_python() {
  python -c '
import shlex, sys
for item in shlex.split(sys.stdin.read()):
    sys.stdout.write(item + "\0")
' <<<"$*"
}

readarray -t example_lines <<'EOF'
'\"'
"\""
'$foo${foo}$(pwd)$((1+2))'
"$foo${foo}$(pwd)$((1+2))"  
EOF

for line in "${example_lines[@]}"; do
  printf 'Input line:         %s\n' "$line"
  printf 'Parsed with xargs:  '; parse_with_xargs "$line" 2>&1; echo
  printf 'Parsed with python: '; parse_with_python "$line" 2>&1; echo
  echo
done

---

Output:

Input line:         '\"'
Parsed with xargs:  \"
Parsed with python: \"

Input line:         "\""
Parsed with xargs:  xargs: unmatched double quote; by default quotes are special to xargs unless you use the -0 option

Parsed with python: "

Input line:         '$foo${foo}$(pwd)$((1+2))'
Parsed with xargs:  $foo${foo}$(pwd)$((1+2))
Parsed with python: $foo${foo}$(pwd)$((1+2))

Input line:         "$foo${foo}$(pwd)$((1+2))"  
Parsed with xargs:  $foo${foo}$(pwd)$((1+2))
Parsed with python: $foo${foo}$(pwd)$((1+2))