From 0c99f0bb022c1181b1199696d0d54e7abd76d0fe Mon Sep 17 00:00:00 2001
From: Vidar Holen <vidar@vidarholen.net>
Date: Sat, 23 Jul 2022 10:05:51 -0700
Subject: [PATCH] Created SC2319 (markdown)

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+## This $? refers to a condition, not a command. Assign to a variable to avoid it being overwritten.
+
+### Problematic code:
+
+```sh
+mycommand
+if [ $? -ne 0 ] && [ $? -ne 14 ]
+then
+  echo "Command failed"
+fi
+```
+
+or
+
+```sh
+mycommand
+[ $? -gt 0 ] && exit $?
+```
+
+### Correct code:
+
+```sh
+mycommand
+ret=$?
+if [ $ret -ne 0 ] && [ $ret -ne 14 ]
+then
+  echo "Command failed"
+fi
+```
+
+or
+
+```sh
+mycommand || exit $?
+```
+
+
+### Rationale:
+
+ShellCheck found a `$?` that always refers to a condition like `[ .. ]`, `[[ .. ]]`, or `test`.
+
+This most commonly happens when trying to inspect `$?` before doing something with it, such as inspecting it again or exiting with it, without realizing that any such inspection will also overwrite `$?`.
+
+In the first problematic example, `[ $? -ne 14 ]` will never be true because it only runs after `[ $? -ne 0 ]` has modified $? to be 0. The solution is to assign `$?` from `mycommand` to a variable so that the variable can be inspected repeatedly.
+
+In the second problematic example, `exit $?` will always `exit 0`, because it only runs if `[ $? -gt 0 ]` returns success and sets `$?` to 0. The solution could again be to assign `$?` to a variable first, or (as shown) use `mycommand || exit $?` as there is no condition to overwrite `$?`.
+
+### Exceptions:
+
+If you intentionally refer to a condition to get its exit status, as in `[ -e file ]; exists=$?`, you can ignore this warning. Alternatively, write it out as in `if ! [ -e file ]; then exists=0; else exists=1; fi`.
+
+### Related resources:
+
+* Help by adding links to BashFAQ, StackOverflow, man pages, POSIX, etc!
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