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BASH - Arithmetic evaluation and errexit trap

count=0
things="0 1 0 0 1"
 
for i in $things;
do
   if [ $i == "1" ]; then
       (( count++ ))
   fi
done
 
echo "Count is ${count}"

Looks fine? I've probably written this many times. There's a small gotcha:

((expression)) The expression is evaluated according to the rules described below under ARITHMETIC EVALUATION. If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to let “expression”.

When you run this script with -e or enable errexit – probably because the script has become too big to be reliable without it – count++ is going to return 0 (post-increment) and per above stop the script. A definite trap to watch out for!