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in reply to Re: array of binaries to hex conversion
in thread array of binaries to hex conversion

Hello jwkrahn

I have just read perlpacktut and I would like to have some clue for my 2 questions about this thread.

What OP does is in short...

printf "0x%x\n", #print it as hex unpack("N", #unpack it as 32bit int ,big endian pack("B32", #pack it as MSB 32bit integer sprintf("%032s", #32 length bit string join('', @rand1))));
1. Does it really MSB ?

MSB should pad '0' for right side, not left side, doesn't it ?
my @rand1 = qw( 1 );
this currenty goes "00000000000000000000000000000001". Is it OK?

2. How oct insures it is MSB and 32bit big endian?

Does oct thinks it as 'long long' seeing the size of $Config{longlongsize}? How it thinks for MSB?
use Config; print "int size=".$Config{intsize}."\n"; print "short size=".$Config{shortsize}."\n"; print "long size=".$Config{longsize}."\n"; print "long long size=".$Config{longlongsize}."\n";
If I could have some explanations by you or other monks, I am glad.

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Re^3: array of binaries to hex conversion
by Anonymous Monk on Sep 12, 2012 at 02:15 UTC

      Thanks for reply. Stories of bit is a little bit hard for me. Please point out if I am saying something wrong.

      How/what?

      I was wrong. I was confused with little endian and big endian. Here OP expects it's input bit array as Big Endian(N) and descending bit order(B).

      printf "0x%x\n",                #hex
          unpack("N",                 #unpack it as 32bit int ,big endian
              pack("B32",             #pack it as MSB descending bit, 32bit inte
                  sprintf("%032s",    #32 length bit string
                      join('',
                          @rand1))));
      
      
      So, left pad works fine, even if it was an array of 16bit.
      my $num;
      $num=pack('n',7); #16bit short Big endian, it's value is 7
      printf "big endian, descending bit order:%s\n", unpack('B16',$num);
      printf "big endian, ascending  bit order:%s\n", unpack('b16',$num);
      printf "unsigned short=%d\n",unpack('n',pack('B16', unpack('B16', $num)));
      
      printf "right pad:%s\n", unpack('B16',$num) . '0' x 16;
      printf "32bit right pad=%d\n",unpack('N',pack('B32', unpack('B16',$num) . '0' x 16));
      #==> this prints "32bit right pad=458752"
      printf "left  pad:%s\n", '0' x 16 . unpack('B16',$num);
      printf "32bit left  pad=%d\n",unpack('N',pack('B32', sprintf('%032d', unpack('B16',$num)))
      #==> this prints "32bit left  pad=7"
      
      You are the same anonymous with this.
      print  '0x'. unpack 'H*', pack("B32", substr("0" x 32 . join('',@rand1), -32));
      
      So, you are going to say: If you need it's hexadecimal values, there is no need for decimal( unpack 'N') -> hexadecimal conversion( printf %x), just print it as hex( unpack 'H').

      I am going to see your links. Thanks for reply.