Bash arithmetic evaluation and errexit traps

Original: https://www.technovelty.org//linux/bash-arithmetic-evaluation-and-errexit-trap.html

In the "Traps for new Players" chapter:

Count=0things="0 1 0 0 1" forI in$things; Doif [ $i == "1" ]; Then       ((count++))   fi DoneEcho "Count is ${count}"

Looks very normal?

I may have written it so many times.

But this is an unexpected error:

(expression)

expression is evaluated according to the rules described by arithmetic EVALUATION. The value of the expression is assumed to be not 0. The return value is 0; Otherwise the return value is 1. This is the same as let "expression".

When you use - E or enable Errexit to run the script-perhaps because the script is too large to be trusted- count++ will return 0 ( post-increment) then the script exits. This trap needs attention!

Bash arithmetic evaluation and errexit traps