Bash arithmetic evaluation and errexit traps

Bash arithmetic evaluation and errexit traps

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

In Chapter traps for new players:

Count = 0 things = "0 1 0 0 1" for I in $ things; do if [$ I = "1"]; then (count ++ )) fidoneecho "Count is $ {count }"

Does it look normal? I may have written this many times. However, this is an unexpected error:

(Expression ))The expression is evaluated according to the rules described by arithmetic evaluation. If the expression value is 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 errexitWhen executing the script -- maybe the script is too large to be untrusted --Count ++Returns0(Post-increment) then the script exits. This trap needs to be noted!