Re: set -e again

[Crystal Kolipe <kolipe.c%exoticsilicon.com@localhost>]

On Wed, Jan 14, 2026 at 01:18:47AM +0000, David Holland wrote:
> I discovered the following today.
> 
> This:
>    set -e
>    f() {
>        echo foo
>        false
>        echo bar
>    }
>    f
> 
> as one might expect, prints "foo" and exits.
> 
> However, this:
>    set -e
>    f() {
>        echo foo
>        false
>        echo bar
>    }
>    f || echo baz
>    echo buzz
> 
> prints "foo", "bar", "buzz", and continues. Furthermore, all the
> shells I have in easy reach agree on it.
> 
> This seems wrong - the exit status of the f is guarded, but the false
> is not. But also, the fact that everybody agrees makes me think it's
> probably the agreed result of the last round of POSIX wrangling over
> the -e definition some years back.
> 
> Is this intended, and is there a way to get the user's intended
> behavior of exit on unchecked failure back? (E.g. is there a different
> set to get functions to return on unchecked failure that might cover
> this?)

Does this achieve what you want?

#!/bin/sh
set -e
f() {
	echo foo
	false
	echo bar
}
f
[ $? ] || echo baz
echo buzz