Re: variables gets shared to child but resets back after exiting fork
hey,
thanks for the reply. The "-|" was the right option for me. The reason is
because I only need to get the first 4 lines of output of the external program.
"-|" was perfect because I don't have to base the stopping on time but rather
on the number of output lines. Another problem solved!!!
Till next time!
--- On Wed, 5/27/09, Chas. Owens wrote:
> From: Chas. Owens
> Subject: Re: variables gets shared to child but resets back after exiting fork
> To: "Michael Alipio"
> Cc: "begginers perl.org"
> Date: Wednesday, May 27, 2009, 7:45 PM
> On Wed, May 27, 2009 at 07:24,
> Michael Alipio
> wrote:
> >
> > Sorry about the indention... must be the mail client
> i'm using.
> > I need to fork because the external program I want to
> run inside the child runs infinitely. I want to have a timer
> running in the parent and after that, kill the child.
> Without forking, I have to do a pkill myprogname
> > I'm reading the perlipc now.. nothing so far..
> snip
>
>
> You don't need a fork for that, you need alarm[1] and block
> eval[2]:
>
> eval {
> local $SIG{ALRM} = sub { die "timeout\n" };
> alarm 3; #die after three seconds
> #operation that needs to die if it is not
> finished in three seconds;
> alarm 0; #turn off the alarm clock
> 1; #make the eval return true
> } or {
> #propagate the error unless it is the
> timeout
> die $@ unless $@ eq "alarm\n";
> };
>
>
>
> 1. http://perldoc.perl.org/functions/alarm.html
> 2. http://perldoc.perl.org/functions/eval.html
>
> --
> Chas. Owens
> wonkden.net
> The most important skill a programmer can have is the
> ability to read.
>
> --
> To unsubscribe, e-mail: beginners-unsubscr...@perl.org
> For additional commands, e-mail: beginners-h...@perl.org
> http://learn.perl.org/
>
>
>
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Re: variables gets shared to child but resets back after exiting fork
On Wed, May 27, 2009 at 07:24, Michael Alipio wrote:
>
> Sorry about the indention... must be the mail client i'm using.
> I need to fork because the external program I want to run inside the child
> runs infinitely. I want to have a timer running in the parent and after that,
> kill the child. Without forking, I have to do a pkill myprogname
> I'm reading the perlipc now.. nothing so far..
snip
You don't need a fork for that, you need alarm[1] and block eval[2]:
eval {
local $SIG{ALRM} = sub { die "timeout\n" };
alarm 3; #die after three seconds
#operation that needs to die if it is not finished in three seconds;
alarm 0; #turn off the alarm clock
1; #make the eval return true
} or {
#propagate the error unless it is the timeout
die $@ unless $@ eq "alarm\n";
};
1. http://perldoc.perl.org/functions/alarm.html
2. http://perldoc.perl.org/functions/eval.html
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
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Re: variables gets shared to child but resets back after exiting fork
Sorry about the indention... must be the mail client i'm using.
I need to fork because the external program I want to run inside the child runs
infinitely. I want to have a timer running in the parent and after that, kill
the child. Without forking, I have to do a pkill myprogname
I'm reading the perlipc now.. nothing so far..
--- On Wed, 5/27/09, Chas. Owens wrote:
> From: Chas. Owens
> Subject: Re: variables gets shared to child but resets back after exiting fork
> To: "Michael Alipio"
> Cc: "begginers perl.org"
> Date: Wednesday, May 27, 2009, 7:15 PM
> On Wed, May 27, 2009 at 05:42,
> Michael Alipio
> wrote:
> >
> > Hi,
> >
> > I have to run an external program but the program does
> not termination on some conditions, e.g, ping, will not exit
> unless you specify -c or some other circumstances.
> >
> >
> > Now I what I want to do is:
> >
> > my @array;
> > die "Cannot fork myprog" unless (defined my $pid =
> fork)
> > if ($pid==0){
> > open MYPROG, "myprog |" or die "Cant run myprog";
> > my $timeout = 0;
> > while (){
> > exit(0) if $timeout == 3;
> > push @array, $_;
> > sleep 1;
> > $timeout++;
> > }
> >
> > waitpid($pid, 0);
> > print "@array\n";
> >
> >
> > The problem with the code above is that @array goes
> back to its initial state after exiting the child. No
> contents are printed. I even tried references but it didn't
> work as well.
> >
> > If I don't use fork, I the way I would kill the
> process is by doing a call to pkill. With fork, it would be
> much easier with exit. However the output of external
> program gets discarded.
> >
> > Can you think of any workaround for this?
> snip
>
>
> Variables are not shared between parent and child.
> The values of the
> parent's variables are copied to the child at the time of
> the fork.
> You really need to read perldoc perlipc[1]. And you
> need to learn how
> to indent code. Leaving your code all against the
> left side of the
> screen makes it hard to read.
>
> Of course, the biggest question is why are you bothering to
> fork in
> the first place? Why not just say
>
> my $program = "myprog";
>
> open my $pipe, "-|", $program
> or die "Cant run $program: $!";
>
> my @array;
> for (1 .. 3) {
> last unless defined(my $line =
> <$pipe>);
> push @array, $line;
> }
>
> close $pipe;
>
>
>
> 1. http://perldoc.perl.org/perlipc.html
>
> --
> Chas. Owens
> wonkden.net
> The most important skill a programmer can have is the
> ability to read.
>
--
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Re: variables gets shared to child but resets back after exiting fork
On Wed, May 27, 2009 at 05:42, Michael Alipio wrote:
>
> Hi,
>
> I have to run an external program but the program does not termination on
> some conditions, e.g, ping, will not exit unless you specify -c or some other
> circumstances.
>
>
> Now I what I want to do is:
>
> my @array;
> die "Cannot fork myprog" unless (defined my $pid = fork)
> if ($pid==0){
> open MYPROG, "myprog |" or die "Cant run myprog";
> my $timeout = 0;
> while (){
> exit(0) if $timeout == 3;
> push @array, $_;
> sleep 1;
> $timeout++;
> }
>
> waitpid($pid, 0);
> print "@array\n";
>
>
> The problem with the code above is that @array goes back to its initial state
> after exiting the child. No contents are printed. I even tried references but
> it didn't work as well.
>
> If I don't use fork, I the way I would kill the process is by doing a call to
> pkill. With fork, it would be much easier with exit. However the output of
> external program gets discarded.
>
> Can you think of any workaround for this?
snip
Variables are not shared between parent and child. The values of the
parent's variables are copied to the child at the time of the fork.
You really need to read perldoc perlipc[1]. And you need to learn how
to indent code. Leaving your code all against the left side of the
screen makes it hard to read.
Of course, the biggest question is why are you bothering to fork in
the first place? Why not just say
my $program = "myprog";
open my $pipe, "-|", $program
or die "Cant run $program: $!";
my @array;
for (1 .. 3) {
last unless defined(my $line = <$pipe>);
push @array, $line;
}
close $pipe;
1. http://perldoc.perl.org/perlipc.html
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
--
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variables gets shared to child but resets back after exiting fork
Hi,
I have to run an external program but the program does not termination on some
conditions, e.g, ping, will not exit unless you specify -c or some other
circumstances.
Now I what I want to do is:
my @array;
die "Cannot fork myprog" unless (defined my $pid = fork)
if ($pid==0){
open MYPROG, "myprog |" or die "Cant run myprog";
my $timeout = 0;
while (){
exit(0) if $timeout == 3;
push @array, $_;
sleep 1;
$timeout++;
}
waitpid($pid, 0);
print "@array\n";
The problem with the code above is that @array goes back to its initial state
after exiting the child. No contents are printed. I even tried references but
it didn't work as well.
If I don't use fork, I the way I would kill the process is by doing a call to
pkill. With fork, it would be much easier with exit. However the output of
external program gets discarded.
Can you think of any workaround for this?
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