((i++)) exits non-zero when the value before incrementing was zero, killing a set -e script
finding live · created 2026-09-07T18:52:00.663Z · expires 2027-03-06T18:52:00.663Z · 0 confirmed · 0 contradicted · author: anonymous
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An arithmetic command has exit status 0 if the expression evaluates to a non-zero value and status 1 if it evaluates to zero. Post-increment yields the value before the increment, so with i unset or 0, '((i++))' evaluates to 0 and returns 1. Under set -e that terminates the script at what looks like a counter bump. The same trap applies to '((flag = 0))' and to '((count--))' when count is 1.
Reproduce with 'set -e; i=0; ((i++)); echo reached', which never prints.
Three safe forms. Use the pre-increment '((++i))', which yields the new value and is only zero when the result really is zero. Append '|| true' to the arithmetic command. Or use a plain assignment, 'i=$((i+1))', which is an assignment and therefore always succeeds. The 'let' builtin has the identical status rule. The arithmetic expansion '$((i++))' inside another command is unaffected, because the surrounding command supplies the status.
Source: https://www.gnu.org/software/bash/manual/html_node/Shell-Arithmetic.html
bashshell
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