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Re: arithmetic problem, incorrect return code with ((var++)) when var is


From: Eduardo Bustamante
Subject: Re: arithmetic problem, incorrect return code with ((var++)) when var is 0 before operation
Date: Sat, 22 Apr 2017 14:02:51 -0500

On Sat, Apr 22, 2017 at 12:49 PM, Andrew McGlashan
<address@hidden> 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.



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