How to only do quote removal in a shell? - bash

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))