SC2251

This ! is not on a condition and skips errexit. Use && exit 1 instead, or make sure $? is checked.

Problematic code:

set -e
! false

Correct code:

set -e
false && exit 1

Rationale:

ShellCheck has found a command inverted with ! that may have no effect. In particular, it does not appear as a condition in an if statement or while loop, or as the final command in a script or function.

The most common reason for this is thinking that it'll trigger set -e aka errexit if a command succeeds, as in the example. This is not the case: ! will inhibit errexit both on success and failure of the inverted command.

Using && exit will instead exit when failure when the command succeeds.

Exceptions:

ShellCheck will not detect cases where $? is implicitly or explicitly used to check the value afterwards:

set -e;
check_success() { [ $? -eq 0 ] || exit 1; }
! false; check_success
! true; check_success

In this case, you can ignore the warning.

Related resources:

• StackOverflow: Why do I need parenthesis In bash set -e and negated return code