setting result's and operand's default value in bash

Question

bash 4.2

the equation
i=$(( (i+1) %3 ))

I want left i to be 0 when the right i is wrong input like null.

something like i=${((i+3) %3):=0} if I can.

Then you validate `i` first, and then do the calculation and the assignment. — oguz ismail, Jul 27 '21 at 08:09
I was trying to do the validating by setting the default value thus this question — Lunartist, Jul 27 '21 at 08:11
No, I don't think that'll work. You should either declare `i` as an integer beforehand (`declare -i i`) so that it can't be assigned trash, or validate its value using a `case` or `[[` command first. — oguz ismail, Jul 27 '21 at 08:14
@Lunartist: What is a _trash value_? If this can be the case, wouldn't it make more sense to validate `i` against a regexp, and issue an error message, instead of silently skipping it? — user1934428, Jul 27 '21 at 08:51

score 1 · Accepted Answer · answered Jul 27 '21 at 07:44

1

Let's try with:

i=$(( ( ${i-0} + 1) %3 ))

answered Jul 27 '21 at 07:44

Antonio Petricca

it returns `1` on my system – Lunartist Jul 27 '21 at 07:55
Because `1%3` is `1`. – KamilCuk Jul 27 '21 at 08:01
I get it. The title was very misleading. My bad. – Lunartist Jul 27 '21 at 08:29
`${i-0}` handles only cases where `i` is not set. For example `i=,; echo $(( ${i-0} ))` results in `bash: , : syntax error` and `i=k; k=7; echo $(( ${i-0} ))` prints `7` (note that we used `i=k`, *not* `i=$k`. Inside an arithmetic context variables can be full-blown arithmetic expressions, not just numbers) – Socowi Jul 27 '21 at 09:52

1 Answers1