On Fri, 15 Dec 2023 17:21:23 -0400
Giacomo Comes <gco...@gmx.com> wrote:
> Hi,
> I have stumbled upon a bug or something I don't understand while
> using the debug trap.
> Please run the script at the end.
> When debug is turned on, during its execution the program
> prints the line number and the line content which returned a non
> zero value (error).
>
> If you look at the script, the only line which should cause
> a non zero return value is:
> ! :
> However the script shows also a non zero return value
> before executing the 'true' command.
> I can only imagine that the sequence
> if ((0)); then
> before the 'else' is the one causing a non zero
> return value, however the previous:
> if ((0)); then
> :
> fi
> (without the else clause) does not cause a non zero return value.
> Is this the expected behavior (and if yes why)?
Yes, it is to be expected.
$ if false; then true; else echo "$?"; fi
1
> Or is it a bug?
> Seen in bash 4.4 and 5.2.
>
> Giacomo Comes
>
> #!/bin/bash
> debugon () {
> trap 'if (($?)); then echo "$((LINENO-1)): $(sed -n "$((LINENO-1))p"
> "$0")" ; fi' DEBUG
> }
The trap here ought to report LINENO without deducting 1. Otherwise, it is a a
recipe for confusion.
> debugoff () {
> trap '' DEBUG
> }
> debugon
>
> :
> ! :
> if ((1)); then
> :
> fi
> if ((0)); then
> :
> fi
> if ((1)); then
> :
> else
> :
> fi
> if ((0)); then
> :
> else
At this point, the value of $? is 1, prior to executing true - a simple
command. Just as for any other simple command, the trap code shall be executed
beforehand. Consequently, your test observes that $? is arithmetically false
and acts accordingly. Keep in mind that this is the only part of your script in
which an "else" clause is actually reached.
> true
> fi
>
> debugoff
>
>
--
Kerin Millar
