store: Fix: Bash ((var++)) returns exit code 1 with set -e when var is 0

graph/fixes/fix-bash-var-returns-exit-code-1-with-set-e-when-var-is-0-c292af.md

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+---
+id: c292afbb-9b99-4628-8836-7a8d00c797e5
+type: fix
+title: "Fix: Bash ((var++)) returns exit code 1 with set -e when var is 0"
+tags: [bash, scripting, debugging, fix, set-e, arithmetic, dotfiles]
+importance: 0.6
+confidence: 0.8
+created: "2026-02-27T05:47:56.914678+00:00"
+updated: "2026-02-27T05:47:56.914678+00:00"
+---
+
+# Bash Arithmetic Pitfall: ((var++)) with set -e
+
+## Problem
+In bash scripts using `set -e` (errexit), the expression `((var++))` **exits with code 1** when `var` is 0, because the arithmetic expression evaluates to 0 (falsy) before the increment takes effect. This causes the script to terminate unexpectedly.
+
+## Root Cause
+`(( expr ))` uses the arithmetic evaluation exit code: exits 0 if the result is non-zero (truthy), exits 1 if the result is zero (falsy). When `var=0`, `((var++))` evaluates the current value (0) first, returns exit 1, then increments — but `set -e` already killed the script.
+
+## Fix
+```bash
+# WRONG (fails when var == 0 with set -e)
+((var++))
+
+# CORRECT
+var=$((var + 1))
+```
+
+## Alternative
+```bash
+# Also acceptable
+((var += 1)) || true   # suppress exit code, but less readable
+```
+
+## Context
+Discovered while writing dotfiles install.sh/uninstall.sh scripts that used counters initialised to 0.