Perl range operator on Scala Context

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# file:
#   range_operator.pl
# description:
#   this file is the test file to show how to use the range operator in PERL
#
# conclusion:
#   the state transition of the .. operator is as such
#
#
# (0, 0)   false    <----  4  ------ (false)  (1, 1)
#         |                        ^
#               |                        |
#(false -> true)1   Left    ..   right   3 (false -> true)
#               |                        |
#               V                        |
#    (1, 0)   (true)    -----   2 -----> true    (0, 1)
#
#
# left is not evaluate in (true) state
# right is not evaluate in (false) state
#
# state flip happen on next round.
use strict;
my $line_no = 0;
print "start", "\n";
my $i = 100;
# the combination of (0, 1)
# and (1, 0) will evaluate as true
#
# while (0 .. 1) {
# print ++$i, "\n";
# sleep 1;
# }
open file, "< range_operator.pl";

# as you can see from the below output,
# the line 1 -- 6 return false on expression 1 . . /^$/
#  
my $i = 0;
$i = 0;
my $line_end = 0;
line:
while (<file>) {
print ("\n") if !$line_end;
print "iteration -- cut point ", ++$i, " | ";
$line_end = 0;
next line if (1 .. /^$/);
$line_end = 1;
print "iteration", $i, "\n";
}
open file, "< range_operator.pl";

print "done", "\n";
 　　The explaination:
　　1.  while (0 .. 1) return true because  (0, 1) on the state chart is true; so you will see a line  printed every second.
　　2. The above output shall be as follow.
　　iteration -- cut point 1 |
　　...
　　iteration -- cut point 23 |
　　iteration -- cut point 24 | iteration24
　　...
　　as you can see, 

22: # state flip happen on next round.
23:
24: use strict;
25:
 　　though at line 23, the expression /^$/ shall evalute as true, according to the state map, the range operator shall be false, and according to the logic at line 23, it shall be the following output.
　　iteration -- cut point 23 | iteration23
　　But why?
　　It is because the state transition will happen the next round. So that explains why the output changes on line 24 rather than line 23;
　　Another thing to notice is that, after the line 24, the output does not change no matter what the text changes to... The reason is because:
　　

• right is not evaluate in (false) state
  
• the left is a constant value (1)