Is "exit ${?}" correct to use in bash?

The [ command in your if statement will set $? after it checks. You'll need to save the original exit status before testing.

some_command
es=$?
if [ "$es" -ne 0 ]; then
    exit "$es"
fi

The syntax is correct, but $? is reset by the [ ... ] command in the if statement. By definition, if the if block is entered then the [ ... ] test must have been successful, and $? is guaranteed to be 0.

You'll need to save it in a variable.

result=$?
if ((result != 0)); then 
    exit "$result"
fi

Alternately, it's more idiomatic to test the result of a command directly rather than testing $?. If you do so then you don't have the problem of $? changing.

if command; then
    echo success
else
    exit   # `exit` is equivalent to `exit $?`
fi

If you don't care about success then you can use ||:

command || exit

If you want to exit on error and preserve the EXIT code of the command, you can enable the errexit option:

• Either with: set -e
• Either with: set -o errexit

See: help set | grep -F -- -e

-e Exit immediately if a command exits with a non-zero status.

errexit same as -e

Alternatively you can trap the ERR signal and use this to exit with the return code of the error.

This will save you from dealing with the -e option consequences.

#!/usr/bin/env bash

err_handler() {
  set -- $?
  printf 'The error handler caught code #%d\n' "$1" >&2
  exit "$1"
}

trap 'err_handler' ERR

create_error() {
  [ $# -ne 1 ] && return 0
  printf 'Create error #%d\n' "$1"
  return "$1"
}

create_error "$@"

Testing with different errors:

for e in 1 0 2; do ./a.sh "$e" ;echo $?; done
Create error #1
The error handler caught code #1
1
Create error #0
0
Create error #2
The error handler caught code #2
2