RE: increment & decrement error when variable is 0

[Jetzer, Bill]

Thank you for the clarification.  I understand now.

--bill

From: Ilkka Virta <itvirta@gmail.com>
Sent: Tuesday, November 24, 2020 2:08 AM
To: Jetzer, Bill <jetzerb@svaconsulting.com>
Cc: bug-bash@gnu.org
Subject: Re: increment & decrement error when variable is 0

On Tue, Nov 24, 2020 at 12:07 AM Jetzer, Bill 
<jetzerb@svaconsulting.com<mailto:jetzerb@svaconsulting.com>> wrote:
                                ((--x)) || echo "err code $? on --x going from 
$i to $x";

err code 1 on ++x going from -1 to 0

That's not about --x, but of the ((...)) construct:

"" (( expression ))
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" See Bash Builtins, for a full description of the let builtin.""

https://www.gnu.org/software/bash/manual/html_node/Conditional-Constructs.html#Conditional-Constructs
Note the second sentence and try e.g.:

$ if (( 100-100 )); then echo true; else echo false; fi
false

That also matches how truth values work in e.g. C, where you could write if 
(foo) { ... } to test if foo is nonzero.
Also, consider the return values of the comparison operators. The behaviour 
here makes it possible to implement
them by just returning a number, instead of having to deal with a distinct 
boolean type that would affect the exit status:

$ (( a = 0 < 1 )); echo $a
1

> err code 1 on x++ going from 0 to 1

As to why you get this here, when going to one instead of zero, remember the 
post-increment returns the original value.
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