While reading the cookbook, I started playing with the forking open. I have a small program (below) that forks 3 processes, each of which modify STDIN and send to STDOUT. What I don't understand is why they work in the order that they do. 
tee('foo');
tee('bar');
tee('baz');

while (<>) { print }

sub tee {

    my $key = shift;

    # reopen STDOUT in parent and return
    return if my $pid = open(STDOUT, "|-");
    die "cannot fork: $!" unless defined $pid;

$|++;

    # process STDIN in child
    while (<STDIN>) {
            print "$key: $_" or die "tee output failed: $!";
    }
    print "$$ exiting...\n";
    exit;  # don't let the child return to main!
}

[download]
When the first child is forked, the STDOUT of the parent is connected to the STDIN of the child, like so: 
Parent -> Child1
For the second and third forks I would expect the new child to be inserted between the parent and current child, so that after 3 forks: 
Parent -> Child3 -> Child2 - Child1
So I would expected each line of output to be preprended with: 
baz: bar: foo:
Instead, each line of output is prepended with: 
foo: bar: baz:
What is the error in my reasoning? How does this actually work? Thanks.

In reply to Understanding forking open by Anonymous Monk

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