What does ${!i} do in bash? In that context, what is the effect of ((i+=1))?

Question

I do not find anything about the meaning of this:

case ${!i} in
     --fa)
       ((i+=1))
       fa=${!i}
       ;;

What is the meaning of

Answer 1

${!i} refers to the variable whose name is the value of $i. It is called variable indirection. See an example to make it more clear:

$ i="hello"      # variable $i contains 'hello'
$ hello="bye"    # variable $hello contains 'bye'
$ echo "${!i}"   # when doing variable expansion of $i, it fetches $hello
bye

Regarding ((i+=1)), it is a way to increment the variable i. See:

$ i=3
$ ((i+=1))
$ echo $i
4

You can also use either of these:

i=$((i+1))

((i++))

let "i=i+1"

---

For further reference, see Bash Reference Manual - Shell Parameter Expansion:

If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion