On Sat, Apr 22, 2017 at 12:49 PM, Andrew McGlashan <andrew.mcglas...@affinityvision.com.au> wrote: [...] > > The return code from ((i++)) operation is different when i has an > initial value of 0.
This is not a bug. Please read the Bash reference manual section on conditional constructs: https://www.gnu.org/software/bash/manual/bash.html#index-commands_002c-conditional | The arithmetic expression is evaluated according to the rules described below (see Shell Arithmetic). 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" This is also described in other sources: 1. http://wiki.bash-hackers.org/commands/builtin/let#description 2. http://mywiki.wooledge.org/ArithmeticExpression#Arithmetic_Commands Also see the 1st example in http://mywiki.wooledge.org/BashFAQ/105 ("Why doesn't set -e (or set -o errexit, or trap ERR) do what I expected?"), which mentions this pitfall. You can also find extensive discussions in the bug-bash archives on why you shouldn't be using set -e, regardless of how many "sources" on the web claim that it helps write "robust" or "correct" shell scripts.
