On Jun 16, 3:00 pm, rvtol+use...@isolution.nl ("Dr.Ruud") wrote:
> On 2011-06-16 19:16, C.DeRykus wrote:
>
>
>
> > Ruud:
> >> C.DeRykus:
> >>> Another solution, not necessarily more elegant, but
> >>> more familiar to most is an eval {} and alarm pair:
>
> >>> EVAL: {
> >>> eval {
> >>> local $SIG{ ALRM } = sub { die "alarm"; };
> >>> local $SIG{ USR1 } = sub { die "usr1" };
> >>> alarm $sleeptime;
> >>> ... some long running operation here ....
> >>> alarm 0;
> >>> 1;
> >>> } or do {
>
> >> Insert:
> >> my $eval_error = $@ || "Zombie error!";
>
> > Huh? If you insert that statement before doing 'alarm 0',
> > then, there's a potential have a race condition with the
> > alarm going off and terminating the entire program with
> > "alarm clock" before you can even check $@.
>
> Realize that the alarm is then already reset (to the earlier setting)
> because the localization gets out of scope.
No, that's not true. The localized signal handler gets
reset but any alarm that is started in that scope will
still be delivered to the current signal handler unless
turned off.
>
> > eval { ... ; "foo" happens and sets $@='bar';
> > ...
> > };
> > my $eval_error = $@ || "Zombie error!";
> > # alarm actually does go off now
> > # and terminates program
> > alarm 0; # too late
> > ....
> > Therefore you wouldn't want to insert any statement
> > before turning off the alarm.
>
> Because the alarm signal is localized (to the eval{}), it is not active
> in the do{}.
No, it's still active. In the case below, the alarm
that was set inside the scope gets delivered to
the default alarm handler. On freebsd 8.2 for
instance:
$ perl -e 'eval{ local $SIG{ALRM}=sub{die "foo"};
alarm 1}; <>'
Alarm clock: 14
So the alarm that was launched in the eval {} scope
is still active and, even though the localized handler
goes out of scope, the delivery still occurs later to
the default handler.
> ...
--
Charles DeRykus
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