in reply to Re: Testing & Using Bits
in thread Testing & Using Bits

Your code works here:

my $bytes = 0x123456;
print join(" ", map { sprintf "%#02x", $_ } unpack("C*",pack("L",0x123
+45678))), "\n";
print join(" ", map { sprintf "%#02x", $_ } unpack("C*",pack("L",$byte
+s))), "\n";

#output:
#0x78 0x56 0x34 0x12
#0x56 0x34 0x12 00

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What did you expect it to do?

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Re^3: Testing & Using Bits
by Aim9b (Monk) on May 31, 2007 at 18:34 UTC
    That's exactly what I expected. I'll recheck my code, but I was getting all zeroes. I even cut/pasted the line & THEN substituted the scalar for the string. Also where can I fine the doc on the sprintf codes %#02x, etc. Thanks.
[reply]
      %#02x means:

      • % A format code
      • # prefix with "0x"
      • 02 2 digits wide, padded with a leading zero if necessary
      • x hexadecimal

      You can read the full documentation here.

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      OK, I guess there's just no getting around it... ;-) Here's my code... 
       # !perl      DebugBCD.pl
open (inFILE, "<C:/Dev/Perl_Dev/ItemExp/TESTFILE.DAT") || 
                    die "Cannot open file\n";

# The first 6 bytes of my test file contain the 12 digit UPC Barcode 1
+23456789910

binmode (inFILE);
seek (inFILE,0,0);
read (inFILE,$inBuf,6);
print "Len=".length($inBuf)."  Raw Data = $inBuf \n";  # This prints t
+he graphics 

$Test = DecodeBCD($inBuf);
print $Test;                    # This prints   48 48 52 48 I expected
+ 1 2 3 4

close inFile;

sub DecodeBCD {
   my ($i,@lo,@hi,$Byte,$Bits,@pairs);
   my $Hi,$Low;
   my $LoMask = 15;
   my $HiMask = 240;
   my $Zone = 0x30;
   for ($i = 0; $i < 2; $i++) 
   {
      $Byte = substr($_[0],$i,1);
      $lo[$i] = ($Byte & $LoMask) | $Zone;        # These OR statement
+s seem to
      $hi[$i] = (($Byte & $HiMask) >> 4) | $Zone; # result in JUST the
+ $Zone piece,
      $pairs[$i] = join("",$lo[$i],$hi[$i]);      # Implying that the 
+& result is Zero.
      print "\nLo $i=".$lo[$i]."    Hi $i=".$hi[$i]."     $pairs[$i]\n
+";
   }  
   print "(@pairs)\n";
   chomp ($Dummy = <STDIN>);
   join('',@pairs);
}

      [download]
      Please try to keep the laughter to a dull roar.
[reply]
[d/l]
        After

              $Byte = substr($_[0],$i,1);

        [download]

        $Byte will contain a single character, but you want the ASCII code of that character. So, try:

              $Byte = ord substr($_[0],$i,1);

        [download]

        I think:)

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[d/l]
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