From bdaa27c9fdd293ca4e4997303e8410ee3d4cb977 Mon Sep 17 00:00:00 2001
From: koalaman <vidar@vidarholen.net>
Date: Sun, 14 Jun 2015 17:19:03 -0700
Subject: [PATCH] Updated SC2082 (markdown)

---
 SC2082.md | 27 +++++++++++++++++++++++++--
 1 file changed, 25 insertions(+), 2 deletions(-)

diff --git a/SC2082.md b/SC2082.md
index 1a6fe03..ffb294b 100644
--- a/SC2082.md
+++ b/SC2082.md
@@ -8,16 +8,39 @@
 
 ### Correct code:
 
+Bash/ksh:
+
+    # Use arrays instead of dynamic names
+    declare -a var
+    var[1]="hello world"
+    n=1
+    echo "${var[n]}"
+
+or
+
+    # Expand variable names dynamically
     var_1="hello world"
     n=1
     name="var_$n"
     echo "${!name}"
 
+POSIX sh:
+
+    # Expand dynamically with eval
+    var_1="hello world"
+    n=1
+    eval "tmp=\$var_$n"
+    echo "${tmp}"
+
 ### Rationale:
 
-You can expand a variable `var_1` with `${var_1}`, but you can not generate the string `var_1` with an embedded expansion, like `${var_$n}`. Instead, you have to use an indirect reference.
+You can expand a variable `var_1` with `${var_1}`, but you can not generate the string `var_1` with an embedded expansion, like `${var_$n}`. 
 
-You do this by creating a variable containing the variable name you want, e.g. `myvar="var_$n"` and then expanding it indirectly with `${!myvar}`. This will give the contents of the variable `var_1`. 
+Instead, if at all possible, you should use an array. Bash and ksh support both numerical and associative arrays, and an example is shown above. 
+
+If you can't use arrays, you can indirectly reference variables by creating a temporary variable with its name, e.g. `myvar="var_$n"` and then expanding it indirectly with `${!myvar}`. This will give the contents of the variable `var_1`. 
+
+If using POSIX sh, where neither arrays nor `${!var}` is available, `eval` can be used. You must be careful in sanitizing the data used to construct the variable name to avoid arbitrary code execution.
 
 ### Exceptions: