cyadav has asked for the wisdom of the Perl Monks concerning the following question:

Here's the code 

#!/bin/perl -w

my $pid = fork();
if ($pid) {
        print "PARENT --" . "\n";
        waitpid(-1, 0) ;
        print "-- PARENT " . "\n";
} elsif ( $pid == 0 ) {
        print "CHILD --" . "\n";
}
print "PID - $pid , MAIN PRG --" . "\n";

[download]

Child is able to execute this line of code  print "PID - $pid , MAIN PRG --" . "\n"; - that it shouldn't be.

Here's the output - 

PARENT --
CHILD --
PID - 0 , MAIN PRG --
-- PARENT
PID - 53784 , MAIN PRG --

[download]
Can you please explain ? Is something wrong here ? how to solve this ?

Replies are listed 'Best First'.
Re: fork - child executes code outside of block
by hippo (Archbishop) on Mar 28, 2018 at 14:13 UTC

    Your assertion (assumption?) that the child should not be able to execute that print line is false. Why should it not when it is outside the entire "if" block?

[reply]

      Ah! thanks :)

[reply]
Re: fork - child executes code outside of block
by tybalt89 (Monsignor) on Mar 28, 2018 at 14:17 UTC 
    #!/bin/perl -w

my $pid = fork();
if ($pid) {
        print "PARENT --" . "\n";
        waitpid(-1, 0) ;
        print "-- PARENT " . "\n";
} elsif ( $pid == 0 ) {
        print "CHILD --" . "\n";
        exit;
}
print "PID - $pid , MAIN PRG --" . "\n";

    [download]
[reply]
[d/l]

      Why split the parent's code like that? 

      defined( my $pid = fork() )
    or die("fork: $!");

if (!$pid) {
    print "CHILD --" . "\n";
    exit;
}

print "PARENT --" . "\n";
waitpid($pid, 0) ;
print "-- PARENT " . "\n";
print "PID - $pid , MAIN PRG --" . "\n";

      [download]
[reply]
[d/l]

        I just made the minimum change to fix the problem.

        It's not my favorite way of doing fork, which is: 

        if( my $pid = fork )
  {
  # parent
  }
elsif( defined $pid )
  {
  # child
  exit;
  }
else
  {
  die "$! on fork";
  }

        [download]
[reply]
[d/l]

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