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+## This assignment is used again in this `declare`, but won't have taken effect. Use two `declare`s.
+
+(or `local`, `typeset`, `readonly`, `export`)
+
+### Problematic code:
+
+```sh
+declare -i first=$1 current=$first
+```
+
+### Correct code:
+
+```sh
+declare -i first=$1
+declare -i current=$first
+```
+
+### Rationale:
+
+When assigning variables via a command, such as `declare`, `typeset`, `local` etc, the expansion of all arguments happen before all assignments. This means that you can't have a variable assigned and then referenced in the same command.
+
+In the example, if `$1` is 42, the arguments will first be expanded in the current environment into `-i first=42 current=`. They will then be passed to `declare` which will perform the assignments.
+
+To correctly set `current=$first` so that it uses the new value of `first`, use two separate commands as shown.
+
+Note that this only applies when assigning via commands, because arguments are always expanded before commands are invoked. If assigning without a command, as in `first=$1 current=$first`, it will work as expected.
+
+### Exceptions:
+
+If you want to reference the value as it existed before the command, e.g. if swapping variables with `declare x=$y y=$x`, you can ignore this message. However, consider rewriting it anyways for the benefit of any humans reading the code.
+
+### Related resources:
+
+* Help by adding links to BashFAQ, StackOverflow, man pages, POSIX, etc!
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