On 13/02/2024 21:04, Thorsten Glaser via austin-group-l at The Open
Group wrote:
Chet Ramey wrote, according to the list archive
ontinue' is a builtin; continue has a return status; !' says to
negate it. It seems easy to come to the conclusion that the script
should return 1.
Yes, I can see how people could come to the conclusion.
But “continue”, if it unwinds the loop, doesn’t really return,
so its return status does not pass to the ! construct but is
directly set before the next loop round or exit, in mksh (and
other pdsh-derived shells).
That's an implementation detail that is not part of the spec. The spec
says that $? reports "the decimal exit status of the most recent
pipeline". The most recent pipeline is "! continue". You're interpreting
"! continue" as something like "continue; (exit $(( !$? )))" so that
"continue" is a pipeline of its own, but that's neither how it's
specified nor how other shells implement it.
You write:
> After all, the continue utility doesn't know it's called by the !
construct.
In ash-derived shells, this basically works because internally, the
continue utility, like all utilities, does know it's called by the !
construct, because the call is wrapped in a function that checks if the
result was negated. But there are other ways to make this work too.
Cheers,
Harald van Dijk
