Calling alarm more than once

[iulukus]

Hi,
Is it possible to call the alarm more than once. If it is posible,
how?
I couldn't do it in my program:

{
......
......
RETRYrint "Send REQUEST to $server\n";

$payload = "x87\x00\x0c\x02\x1b\x62\x1f\x73\x96\x40\x68\x74\xfb\xff
\xff";

$lpayload = length($payload);
$dummy = syswrite($remote, $payload, $lpayload); # $remote is my UDP
socket.
print "Send REQUEST \n";

eval {
local $SIG{'ALRM'} = sub {print "abc";die "alarm\n"};
alarm 6;
sysread($remote, $result, 1024); # Read from socket
alarm 0;
};
if ($@) {
die unless $@ eq "alarm\n"; # propagate unexpected errors
goto RETRY;
# timed out
}
else {
print $@;
# didn't
}

OUTPUT:
Send REQUEST to 1.1.1.1
Send REQUEST
abcalarm
Send REQUEST to 1.1.1.1
Send REQUEST


For the second loop, program did not call the alarm subroutine and
program hang.
What can be the problem?
thanks in advance.

 

[Tim Southerwood]

Hi,
Is it possible to call the alarm more than once. If it is posible,
how?

Assuming you are doing this on a typical unix system...

No, unfortunately:

man -S2 alarm

....
DESCRIPTION
alarm() arranges for a SIGALRM signal to be delivered to the process
in seconds seconds.

If seconds is zero, no new alarm() is scheduled.

In any event any previously set alarm() is cancelled.
....

I couldn't do it in my program:

No, but that's not an error. PITA, but expected.

I didn't have time to read your code, but in order to run multiple pending
alarms, one solution is to write your own multi-alarm
subroutine/module/class, which queues alarm requests. That class then uses
alarm() to time out to the next alarm scheduled to trip and does a callback
to a user supplied subroutine, after setting up the next alarm().

Beware of sleep() - that uses the same timer as alarm() so you may need to
emulate that. Also beware of random other modules that use alarm() or
sleep() internally.

As usual, someone already had this problem before you and CPAN to the
rescue:

http://search.cpan.org/~johnsca/libalarm-1.0/lib/Alarm/Queued.pm

Is that any use to you?

HTH

Tim

 

[Peter Scott]

Assuming you are doing this on a typical unix system...

No, unfortunately:

man -S2 alarm

Or on any system that Perl implements alarm():

perldoc -f alarm
[...]
Only one timer may be counting at once. Each call disables the
previous timer, and an argument of 0 may be supplied to cancel
the previous timer without starting a new one. The returned
value is the amount of time remaining on the previous timer.

 

[xhoster]

Hi,
Is it possible to call the alarm more than once.

Yes, but you can't have them nested.

If it is posible,
how?

Just do it.

I couldn't do it in my program:

{
.....
.....
RETRYrint "Send REQUEST to $server\n";

$payload = "x87\x00\x0c\x02\x1b\x62\x1f\x73\x96\x40\x68\x74\xfb\xff
\xff";

$lpayload = length($payload);
$dummy = syswrite($remote, $payload, $lpayload); # $remote is my UDP
socket.

I don't have access to your UDP socket. So I took that part of your
code out, and turned the sysread into a sleep 20. It worked as I expected,
retrying repeatedly.

print "Send REQUEST \n";

eval {
local $SIG{'ALRM'} = sub {print "abc";die "alarm\n"};
alarm 6;
sysread($remote, $result, 1024); # Read from socket
alarm 0;
};
if ($@) {
die unless $@ eq "alarm\n"; # propagate unexpected errors
goto RETRY;
# timed out
}
else {
print $@;

What is the point of printing $@ when $@ is false?

Perhaps the die-eval is screwing up your socket. If you include an
simplified example for your UDP server, I'd give it a try and see what
happens on my system.

Xho