my $pid = fork;
die unless defined $pid;

if ($pid) {
  # Parent. Do whatever you want
} else {
  # Child
  do 'blastall.pl';
}
[download]

UPDATE: Ok, got it fixed. Shoud be $kid = waitpid(-1,0). I quite don't understand how it matches the correct PID, but this seems to be working.

I'm having problems with the execution and fork. Probably as I have never used it before. So is this correct way to use it? Where does the waitpid go and how does it work? How the program flow go from here?

#!/usr/bin/perl
#
# Wrapper takes is the arguments of the blastall binary
#  changes directory to binarys directory
#  adds binary name
#  runs it
# Needs to be able to trap normal signals and definedly kill the proce
+ss, 
# as it ignores some which it should not.
#
use warnings;
use strict;
use File::Basename;
use sigtrap qw(handler catch_signal normal-signals);
my $pid;        # blastalls pid after fork

sub catch_signal {
  print STDERR "SIGNAL:",shift," blast should be $pid\n";
  kill (9,$pid);
  exit;
}

print "Going\n";

my $dir=dirname($0);
chdir $dir;
unshift (@ARGV,"./blastall");

$pid = fork;
die "Can't fork: $!\n" unless defined $pid;

if ($pid) {
  # Parent. Do whatever you want
  print "In parent.\n";
  my $kid=0;
  do {
    waitpid ($pid,0);
  } until ($kid == -1);
} else {
  # Child
  print "In child.\n";
  exec (@ARGV) or die "Unable to exec: $!";
}

print "After fork if.\n";
# this was used after `@ARGV` to get error message if executing failed
# if ($?!=0) {
#  warn "@ARGV failed: ",$?,":",$?>>8,"\n";
#  exit $?;
# }

print "Exiting.\n";
exit;
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I was kind of hoping that it would not involve fork :).

In fact, you are anyway. Backticks does not avoid it, it only does hide it from you...


--
zejames

note: this pid is actually the pid of the subshell. (updated)

$pid = open( my $proc, "-|", "ls -l" );

print "pid $pid\n";


my @lines = <$proc>;
print @lines;
[download]

No, it's not the pid of the subshell, because there isn't a subshell. Perl only calls a subshell if the command to call contains characters that are special to the shell (except whitespace) - otherwise Perl will split given string on whitespace, and execute the command (with any arguments) directly. In this case, perl will do an execv of ls, passing it -l as its sole argument.

Had you written:

$pid = open(my $proc, "-","ls -l>/tmp/foo");
[download]

then $pid would contain the pid of a shell.