Shell exit silently at arithmetic expression in for loop

A simple example should explain why:

$ ((success++))
$ echo $?
1

The reason is that any arithmetic operation which produces a numeric value of zero returns 1. I don't know what to say - Bash has gotchas enough for the whole world.

It is the consequence of having -e set. Any command with an exit code of 1 (not zero) will trigger an exit.

This script works fine:

#!/bin/bash
(( success++))
echo "Still going 1 $success"

This doesn't

#!/bin/bash
set -e
(( success++))
echo "Still going 1 $success"

Solutions

The simplest is to remove the set -e line.

If that is not an option, Use this:

(( ++success ))

Other alternatives:

#!/bin/bash

set -e
success=0
success=$(( success+1 ))
echo "still going 1 $success"

success=0
(( success=success+1 ))
echo "still going 2 $success"

success=0
(( success+=1 ))
echo "still going 3 $success"

success=0
(( ++success ))
echo "still going 4 $success"

success=0
(( success++ ))
echo "still going 5 $success"

Only the option number 5 will have an exit code of 1.

Other (more complex solutions for any value of variable a).
The first one uses the (POSIX) colon (:) builtin to make it POSIX compatible.

: $(( a+=1 ))        ; echo "6 $a $?"   ## Valid Posix
   (( a++ )) || true ; echo "7 $a $?"
   (( a++ )) || :    ; echo "8 $a $?"
   (( a++ , 1 ))     ; echo "9 $a $?"
   (( a++ | 1 ))     ; echo "10 $a $?"