Hi,
I’ve got the following issue, and… yesssss I can see how the
reporter could come to the conclusion that it should “return”
1, but…
… at what point does “continue” “return”? Where do I stop
operating?
mksh does indeed deviate from all the other shells here, but
I’m puzzled how they’d even implement that; mksh in the case
of a break or continue unwinds to the enclosing loop, then
sets the variable tracking the exit status to 0; the continue
command does not rightly return if it executes.
If it did, where would we stop? { continue; ls; } && ls || ls
I guess I’m asking for a blessing to support my reading that
continue may unwind as soon as it executes and that writing
“! continue” and expecting the ! to negate the errorlevel if
the continue successfully executes is unreasonable.
---------- Forwarded message ----------
From: Oğuz İsmail Uysal <2053...@bugs.launchpad.net>
Date: Tue, 13 Feb 2024 17:30:06 -0000
Subject: [Bug 2053053] [NEW] for loop exit status
Public bug reported:
The following shell script should print 1 according to POSIX:
for x in y; do
! continue
done
echo $?
But it prints 0 on mksh
** Affects: mksh
Importance: Undecided
Status: New
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https://bugs.launchpad.net/bugs/2053053
Title:
for loop exit status
Status in mksh:
New
Bug description:
The following shell script should print 1 according to POSIX:
for x in y; do
! continue
done
echo $?
But it prints 0 on mksh
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https://bugs.launchpad.net/mksh/+bug/2053053/+subscriptions
