I am learning Perl, and I am trying to see what happens when I write some weird code. And I have some questions...

##################################

sub AA{6}
sub BB{{{3}}}
sub L{AA+BB}  # Now, here Perl thinks that I am
              # calling AA(+BB) so, instead of
              # adding 6+3, it simply returns 6

print "\nL=".L."\n";
print "\nL=".((AA) + BB) ."\n";  # Now I got it right.

# Okay, I understand this. So far everything is clear.

##################################

sub UNIQUE::print{print 'Hello World!'}
UNIQUE->print(333);

sub UNIQUE::print{ print "\nMuhahaha! @_ \n"; }
UNIQUE->print(65.65.65.65);   # Okay this is weird.
# First of all, how does 65 become letter 'A' without using chr(65) ?
# Secondly, Perl allows me to redefine my UNIQUE::print and now
# this second definition overwrites the first one.
# What happened to the first one???
# Why is my 'Hello World' never showing up at all?
##################################

print "What on earth is this? >>> " . sub {5};

my $D = sub { print "What the Heck!"; };  # What am I doing here?
# And why is this allowed in Perl? What's the point of
# creating a noname sub? Okay, so if I wanted to, how
# would I run my noname sub using the handle $D ?

##################################

sub Red
{
  sub LittleRed{'red'};  # sub within a sub
  LittleRed;             # returns 'red'
}
print "\nPerl is getting confused here..." . Red;
print "\n\n";

print LittleRed;    # Wait. This is not supposed to work!
# If I declare a sub within a sub, then the inner sub should
# not be accessible from the outside. Is this right???
# I mean what's the point of declaring a sub within a sub
# if it's the same as declaring it outside the sub?

###################################

sub X{10}
print "\nYES>".X;     # This prints 10. Fine.

sub Y{my$i=0;$i while($i++<10);}print"\nYES>".Y;

# Now, why is this not printing 10??
# When the while loop ends, $i equals 10, so if I just
# say "$i" then it should use that as a return value for
# the sub. No?

###########################

# Another question: Is there a way to
# access previous return values?
# Example:

sub DDD { 0; 1; 2; 3; 4; 5; }

print DDD;

# Is it possible to do something like this
# without using the = assignment operator?:
#
# sub DDD {
#  4+4;
#  5+(previous lines' result);
#  return previous line's result;
# }
#
# So, DDD would return 13.
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