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+## For indirection, use arrays, `declare "var$n=value"`, or (for sh) read/eval
+
+### Problematic code:
+
+```sh
+n=1
+var$n="hello"
+```
+
+### Correct code:
+
+For integer indexing in ksh/bash, consider using an indexed array:
+
+```sh
+n=1
+var[n]="hello"
+echo "${var[n]}"
+```
+
+For string indexing in ksh/bash, use an associative array:
+
+```sh
+typeset -A var
+n="greeting"
+var[$n]="hello"
+echo "${var[$n]}"
+```
+
+If you actually need a variable with the constructed name in bash, use `declare`:
+
+```sh
+n="Foo"
+declare "var$n=42"
+echo "$varFoo"
+```
+
+For `sh`, with single line contents, consider `read`:
+```sh
+n="Foo"
+read -r "var$n" << EOF
+hello
+EOF
+echo "$varFoo"
+```
+
+or with careful escaping, `eval`:
+
+```sh
+n=Foo
+eval "var$n='hello'"
+echo "$varFoo"
+```
+
+### Rationale:
+
+`var$n=value` is not a valid way of assigning to a dynamically created variable name in any shell. Please use one of the other methods to assign to names via expanded strings. [Wooledge BashFaq #6](https://mywiki.wooledge.org/BashFAQ/006) has significantly more information on the subject.
+
+
+### Exceptions:
+
+None
+
+### Related resources:
+
+* [Wooledge BashFaq #6](https://mywiki.wooledge.org/BashFAQ/006): How can I use variable variables (indirect variables, pointers, references) or associative arrays?
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