From fe4f3036f95c8cbae10140b2222479a11d3a82e4 Mon Sep 17 00:00:00 2001
From: Vidar Holen <vidar@vidarholen.net>
Date: Sat, 23 Jul 2022 10:12:04 -0700
Subject: [PATCH] Created SC2320 (markdown)

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 1 file changed, 40 insertions(+)
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+## This $? refers to echo/printf, not a previous command. Assign to variable to avoid it being overwritten.
+
+### Problematic code:
+
+```sh
+mycommand
+echo "Command exited with $?"
+if [ $? -ne 0 ]
+then
+  echo "Failed"
+fi
+```
+
+### Correct code:
+
+```sh
+mycommand
+ret=$?
+echo "Command exited with $ret"
+if [ $ret -ne 0 ]
+then
+  echo "Failed"
+fi
+```
+
+### Rationale:
+
+ShellCheck found a `$?` that always refers to `echo` or `printf`. 
+
+This most commonly happens when trying to show `$?` before doing something with it, without realizing that any such action will also overwrite `$?`.
+
+In the problematic example, `echo "Command exited with $?"` was intended to show the exit code before acting on it, but the act of showing `$?` also overwrote it, so the condition is always false. The solution is to assign `$?` to a variable first, so that it can be used repeatedly.
+
+### Exceptions:
+
+If you intentionally refer to `echo` to get the result of a write, you can ignore this message. Alternatively, write it out as in `if echo $$ > "$pidfile"; then status=0; else status=1; fi`
+
+### Related resources:
+
+* Help by adding links to BashFAQ, StackOverflow, man pages, POSIX, etc!
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