From b9b0bb7584eb80ee5e2c181cd4d0ed62887144a6 Mon Sep 17 00:00:00 2001
From: koalaman <vidar@vidarholen.net>
Date: Wed, 20 Jan 2016 10:55:23 -0800
Subject: [PATCH] Created SC2056 (markdown)

---
 SC2056.md | 41 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 41 insertions(+)
 create mode 100644 SC2056.md

diff --git a/SC2056.md b/SC2056.md
new file mode 100644
index 0000000..4b013db
--- /dev/null
+++ b/SC2056.md
@@ -0,0 +1,41 @@
+## You probably wanted && here
+
+### Problematic code:
+
+```sh
+if  (( $1 != 0 || $1 != 3 ))
+then
+  echo "$1 is not 0 or 3"
+fi
+```
+
+### Correct code:
+
+```sh
+if  (( $1 != 0 && $1 != 3 ))
+then
+  echo "$1 is not 0 or 3"
+fi
+```
+
+### Rationale:
+
+This is not a bash issue, but a simple, common logical mistake applicable to all languages.
+
+`(( $1 != 0 || $1 != 3 ))` is always true:
+
+* If `$1 = 0` then `$1 != 3` is true, so the statement is true.
+* If `$1 = 3` then `$1 != 0` is true, so the statement is true.
+* If `$1 = 42` then `$1 != 0` is true, so the statement is true.
+
+`(( $1 != 0 && $1 != 3 ))` is true only when `$1` is not `0` and not `3`:
+
+* If `$1 = 0`, then `$1 != 3` is false, so the statement is false.
+* If `$1 = 3`, then `$1 != 0` is false, so the statement is false.
+* If `$1 = 42`, then both `$1 != 0` and `$1  != 3` is true, so the statement is true.
+
+This statement is identical to `! (( $1 == 0 || $1 == 3 ))`, which also works correctly.
+
+### Exceptions
+
+None.