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Which bit is set? (inverse of (1<<$value) operator)

by wirito (Acolyte)
on Sep 17, 2012 at 11:04 UTC ( #994008=perlquestion: print w/ replies, xml ) Need Help??
wirito has asked for the wisdom of the Perl Monks concerning the following question:

Hello fellow monks, I'm storing values in a bitwise manner for speed up comparations. I can have 0 or more values in the 0..8 range, so I can store it this way:
$bits |= (1<<$val)
When 1 single bit is set I want to retrive the original $val. So currently I have this implemented:
if( unpack( "%16B*", pack("n", $bits)) == 1 ) { my $ret = 0; while (!($bits & 0x1)) { $ret++; $bits >>= 1; }; return $ret; };
I don't know why, but the while loop seems ugly. Could you enligthen me with a more perlish way to do this? PS: I hope the title is not so much confusing.

Comment on Which bit is set? (inverse of (1<<$value) operator)
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Re: Which bit is set? (inverse of (1<<$value) operator)
by grizzley (Chaplain) on Sep 17, 2012 at 11:17 UTC
    In description of unpack there is following method to count number of bits:
    # Count the bits in a chunk of memory (e.g. a select vector) unpack( '%32b*', $mask );
    Update: sorry, I misunderstood question. If you have string of bits with one bit set, then you could try just $bitstring =~ /1(0*)$/; $originalvalue = length($1)+1;
      Nope, it's a two-byte int, not a strings filled with 0's and 1's.
Re: Which bit is set? (inverse of (1<<$value) operator)
by BrowserUk (Pope) on Sep 17, 2012 at 11:27 UTC

    1. If only 1-bit will ever be set:

      Use a lookup table:

      %lookup = map{ 2**$_ => $_ } 0 .. 63;; $flags = 0;; $flags |= 1<<(rand 64);; print "The flag set is:", $lookup{ $flags };; The flag set is: 12
    2. If multiple flags can be set:

      Some form of loop is necessary to avoid an impossibly large lookup table:

      $flags = int rand( 2**64 );; print for grep $flags & 1 << $_, 0 .. 63;; 49 50 51 54 55 57 58 59 printf "%u\n", $flags;; 1066790161733386240 $t = 0; print $t += 1<<$_ for grep $flags & 1 << $_, 0 .. 63;; 562949953421312 1688849860263936 3940649673949184 21955048183431168 57983845202395136 202099033278251008 490329409429962752 1066790161733386240

    With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
    Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
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      Many thanks!

      The lookup idea is nice and clean. I will use it as I only have 9 different bits.

      However, your second option is the one I was (foolish) looking for. In my code, this could be the line:

      # return the first less significant bit set. return $_ for grep { $bit & 1<<$_ } (0..8)';

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