---
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: "2026-02-27T05:48:38.596981+00:00"
relations:
  - target: 5f31ea7f-9b53-427d-b601-ab98dabaa2e3
    type: RELATED_TO
    direction: incoming
    strength: 0.8
    edge_id: 5f5d8b39-4f75-4099-bcc6-e2aa87ec3d2b
  - target: 66bac9cf-1b1e-42f5-903d-d28a9d5e1e1a
    type: RELATED_TO
    direction: outgoing
    strength: 0.65
    edge_id: 0115d2f7-d73a-404b-ad3c-d44647d5c1b7
---

# 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.