Can you explain what you really want to do? Comparing the data in "one byte" is usally done just by comparing characters (since, aside from utf8, characters in Perl are one byte long). If you want extract a single bit from a byte you'll usually use & (not &&) and a bitmask (eg. 0b00000001 to get the first bit).

-sam

sub isbitset {

   my $bytetocheck = $_[0];
   my $bittocheck = $_[1];

   if (($bytetocheck >> $bittocheck) & 1) {
      return 1;
   } else {
      return 0;
   }

} # End of isbitset function.

##################################################
# I am testing all 8 bits to see which one is set.
##################################################
printf ("\tByte %08b bit 0 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 0));

printf ("\tByte %08b bit 1 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 1));

printf ("\tByte %08b bit 2 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 2));

printf ("\tByte %08b bit 3 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 3));

printf ("\tByte %08b bit 4 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 4));

printf ("\tByte %08b bit 5 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 5));

printf ("\tByte %08b bit 6 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 6));

printf ("\tByte %08b bit 7 => %d\n\n", 
      ord $uh[0], &isbitset((ord $uh[0]), 7));
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FYI, the following all accomplish the same:
if (($bytetocheck >> $bittocheck) & 1)
if ($bytetocheck & (1 << $bittocheck))
if ($bytetocheck & $bit[$bittocheck])
if (vec($bytetocheck, $bittocheck, 1))

And in the interest of eliminating redundancy,

   if (($bytetocheck >> $bittocheck) & 1) {
      return 1;
   } else {
      return 0;
   }
[download]

simplifies to any of the following:
return (($bytetocheck >> $bittocheck) & 1);
return ($bytetocheck & (1 << $bittocheck)) ? 1 : 0;
return ($bytetocheck & $bit[$bittocheck]) ? 1 : 0;
return vec($bytetocheck, $bittocheck, 1);

and the following, if you're only interested in returning a boolean value:
return ($bytetocheck & (1 << $bittocheck));
return ($bytetocheck & $bit[$bittocheck]);

You could simply return the value of ($bytetocheck >> $bittocheck) & 1, since it will return 0 or 1. And it's unnecessary to use the '&' when calling your function.

---

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

Jeff: Done! Thanks, RV

Oh yeah, other than:

   for ($tmp = 0 ; $tmp < 8 ; $tmp++) {
      printf ("\tByte %08b bit $tmp => %d\n\n", 
          ord $uh[0], &isbitset((ord $uh[0]), $tmp));
   }
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