You have written quite a lot of code here. It is very confusing to us because of the great number of 'unusual' ways you implement things.

I think it would help you if you would check some of your assumptions on how Perl code works. When in doubt I start up perl in the debug mode and try out a line or two. Such as:

perl -de0

Loading DB routines from perl5db.pl version 1.25
Editor support available.

Enter h or `h h' for help, or `perldoc perldebug' for more help.

main::(-e:1):   0
  DB<1> x chr(127)
0  "\c?"
  DB<2> x int(chr(127))
0  0
  DB<3> x ord(chr(127))
0  127
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This is me trying out that expression you were confused about. I find out that chr(127) does indeed convert the number into a string character. I then check what int(chr(127)) gives me - zero! Hmmm, doing an int() on a string that doesn't start with a number returns zero. Just to check what bart mentioned I try ord() and see that that is the reverse of what chr() does (which is why he mentioned it).

So I directly tested that part of the problem expression and find out it will _always_ give me zero.

But the array types doesn't always have numbers in it (so that '==' would be appropriate) nor does it always have characters in it (so that 'eq' would be appropriate). Compare these lines:

  @types[++$#types] = chr $temp;
  @types[++$#types] = int(rand(10));
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In the first you force the value to be a character, but in the second you ask for the value as an integer. So neither '==' nor 'eq' are going to always work for the values in @types.