hOURS am Sonntag, 3. Dezember 2006 03:25:
> "D. Bolliger" <[EMAIL PROTECTED]> wrote: hOURS am Donnerstag, 30. November
2006 21:09:
> > Jen Spinney wrote: On 11/20/06, hOURS wrote:
> > > Recently I posed a question on here regarding a program I have that
> > > runs other programs (through require statements and what not). My
> > > problem was that the programs getting run might have syntax errors and
> > > I wanted to skip over each of those and go onto the next one. We
> > > figured out a way to handle that. It turns out however, that these
> > > programs sometimes have an even more troublesome problem: infinite
> > > loops. I knew about this possibility, but figured I would just use the
> > > time function, and if a program was taking to long, skip over it. Yeah,
> > > that wasn't so smart. I can't have the main program check the elapsed
> > > time while the required program is running its infinite loop. Or can I
> > > somehow? Any ideas anybody? Thank you.
> > > Fred Kittelmann
> >
> > Fred,
> > Have you checked out the alarm function? I'm a beginner myself and I
> > had a similar problem earlier today. alarm seemed to do it for me.
> > Good luck!
> >
> > - Jen
> >
> > Thanks Jen,
> > I've checked out alarm as much as I can. My PERL textbook scarcely
> > mentions it. Trying "perldoc -f alarm" was a little more informative,
> > but I still don't understand how to use this. Can anyone explain it to
> > me? Fred
>
> Does the following modified code example from 'perldoc -f alarm' helps?
>
> Dani
>
> #!/usr/bin/perl
> use strict;
> use warnings;
>
> my $timeout=5; # secs
>
> eval {
> # Assign a signal handler subroutine which is invoked in case
> # the alarm signal is sent
> #
> local $SIG{ALRM} = sub { die "alarm\n" }; # NB: \n required
>
> # "send an alarm signal after $timeout seconds!"
> #
> alarm $timeout;
>
> # to test a non-timeout die, uncomment following line:
> #die;
>
> # the problem:
> #
> endless_loop();
>
> # reset alarm timer: "Don't send alarm signal any more"
> #
> alarm 0;
> };
>
> # Check if the code within eval died because of an alarm signal
> # or something else. We check the die message for that.
> #
> if ($@) {
> if ($@ eq "alarm\n") {
> warn "endless_loop() interrupted after timeout\n";
> }
> else {
> warn "code in eval died!\n";
> die;
> }
> }
>
> warn "program continues...\n";
>
> sub endless_loop { {} while 1 }
>
> __END__
> Thanks, I suppose I understand that code example from 'perldoc -f alarm' a
> little better. But much of it remains mysterious. e.g. the very first
> thing within eval. The only brackets I've ever seen with variables are []
> for list elements. What's going on with {}?
eval{} - see perldoc -f eval, it's the eval BLOCK variant.
$SIG{ALRM} - that's the way to retrieve the value with the 'ALRM' key from the
hash %SIG; analogous to retrieve the $i'th value $array[$i] from an array
@array. See perldoc perldata for the perl data types.
NB: The '$' indicates the data type of the *accessed* data ($SIG{ALRM},
$array[$i]), and '[]' versus '{}' shows you if the "aggregated data
structure" from which you take an element is a hash or an array.
There's another usage of {}:
my $x=1;
{ # make a scope block
my $y=2;
}
# $y is no more defined outside the scope block above.
> And what a strange thing to
> set a variable to - seems to be neither string nor number, but a
> subroutine?
You refer to the line
local $SIG{ALRM} = sub { die "alarm\n" };
?
Assigned is a so called "anonymous subroutine".
Above line could be reformulated by
sub alarm_sburoutine { die "alarm\n" };
local $SIG{ALRM} = \&alarm_subroutine;
But since the subroutine is only used once and is very short, the original
line is an abbreviation.
see perldoc perlsub
> And why would you have a subroutine with just one line?
Why not?
sub a_sub_returning_one { 1 };
print 'the sub returns ', a_sub_returning_one;
Of course you could write it like
sub {
die "alarm\n";
}
White space in the program code does not matter in *most* cases, unless in
python for example.
> And
> how can you have a subroutine without a name?
See above, "anonymous subroutine" :-)
> And without a call to it?
It *is* called at the time the alarm timer reaches 0, although not explicitly
from the code;
See perldoc perlipc; it explains signals handlers at the top.
> Where is the thing being timed? I understand something is being given 5
> seconds, but what? Why is the variable $SIG{ALRM} not used again?
Hm... somebody else may explain this much better than I;
alarm() is a form of IPC, using signals, and it is provided by the operating
system. perls alarm() is just an interface to the operating systems alarm
function. On *nix, see
man alarm
man signal
So it's timed "in the OS", given 5 secs, and the subroutine ("signal handler"
in this context) is called from the OS internally.
> Is
> there some significance to the name of that variable?
yes, %SIG is one of the special variables of perl, see the following document
where all of them are described:
perldoc perlvar
> In 'perldoc -f
> alarm' there's mention of a SIGALRM, but I don't know what that is.
Im fairly shure that there it is simply a short way to say "The ALRM signal".
> But I think we can ignore all those questions, because I don't see a need
> to work with this example. I'm just looking for someone to tell me how
> alarm works. A few sentences in English will be fine. No code really need
> be written.
Ouch, I did not read the whole post before answering :-(
For the gory details of how it works on operating system / C level, somebody
else has to take over - sorry
Dani
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