Actually only the first value is correct. That's because pack and unpack work on byte aligned positions, so the second set of three bits is read at position 8 rather than 3. This is visible if your input data contains something else than zeros:

use Data::Dumper;
my $counter = 'a0b0c0d0';
my @x= unpack('(B3)(B3)(B2)(B)(B)(B)(B2)(B3)BBBBB(B11)', pack('H*', $c
+ounter));
print Dumper(\@x);
__DATA__
$VAR1 = [
          '101',
          '101',
          '11',
          '1',
          '',
          '',
          '',
          '',
          '',
          '',
          '',
          '',
          '',
          ''
        ];
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FYI, another way to read bits from a string is to use vec, that makes it possible to read N bits from position P.

FYI, another way to read bits from a string is to use vec, that makes it possible to read N bits from position P.

Yes, but it also has byte/word/doubleword/quadword alignment restrictions.

---

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Oh right, thanks. So that's either pack and unpack or binary masking and shifting operations (maybe combined with vec) on integer values then.

my @x = unpack('(a3)(a3)(a2)(a)(a)(a)(a2)(a3)aaaaa(a11)', unpack 'B*',
+ pack('H*', 'a0000001'));
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Your use of parentheses around some of the Bs in the OPed code and not others suggests that you might want to skip over the BBBBB group. If so:

c:\@Work\Perl\monks>perl -wMstrict -MData::Dump -le
"my $counter = '87654321';
 my @x =
   unpack 'a3 a3 a2 a a a a2 a3 x[a5] a11',
   unpack 'B*',
   pack   'H*', $counter
   ;
 dd \@x;
"
[100, "001", 11, 0, 1, 1, "00", 101, "01100100001"]
[download]

Thank a lot , I understood the mistake and could correct it