Beware, I think this code you gave is wrong: 

$pid=fork;
if ($pid==0)
{try and start new process with exec??}

elsif ($pid)
{do what I want to do with this script??}

else {die because of fork error }

[download]
If fork gives an error, it will execute the first body; the third body is never executed.

The correct way is to use the defined function to test for an error, something like this: 

use warnings;
$pid = fork;
if (!defined($pid)) {
    die "fork error";
} elsif (!$pid) {
#line 6 "child"
    warn "starting";
    for (1 .. 5) {
        sleep 1;
        warn "working hard. really. ", $_*20, "%";
    }
    exit; # or exec
} else {
#line 14 "parent"
    warn "starting the parent";
    for (1 .. 2) {
        sleep 1;
        warn "doing some work";
    }
    warn "waiting for the child to finish (or just collecting it if it
+ has finished";
    $pid == waitpid($pid, 0) or
        die "error waiting: $!";
    0 == $? or
        die "child has died";
    warn "ok, collected the child successfully";
}
__END__

[download]

In reply to Re: Confused by fork examples by ambrus
in thread Confused by fork examples by webshark

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