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

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
[download]

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
[download]

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Re^2: Which bit is set? (inverse of (1<<$value) operator) by wirito (Acolyte) on Sep 17, 2012 at 12:11 UTC
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)';` [download]	[reply] [d/l]