Thank you for all the work you've put into this.

But whilst this finds a match (if it exists), I'm having the same problem with this as I am with my attempts to implement Dave_the_M's suggestion, that of converting the match to a bit position at which it was found?

$-[0] tells me the byte position, but then I need to work the byte at than position back to the pattern than matched and then the bit offset within it. Thoughts?

Put in match groups, i.e., replace

my $p = join '|', map { "$pre[$_]\\0{@{[($N-$_)>>3]}}$suf[($N-$_)&0x7]
+" } 0..7;
[download]

with

my $p = join '|', map { "$pre[$_](\\0{@{[($N-$_)>>3]}}$suf[($N-$_)&0x7
+])" } 0..7;
[download]

By construction, only one will match and $#- will tell you which one. We place the leftparen so that the group always starts at the first byte that's beginning with a zero bit. Then you just have to subtract the number of zero bits at the end of the byte that $pre[$_] is matching, but that's just $#- - 1. So the answer is

$-[$#-]*8 - $#- + 1
[download]

Still needs a bit more work to handle the $N < 8 case.

Unfortunately, the bitpatterns you generate -- in common with those used by grimey's code -- are not compatible with vec.

By way of example, this generates a vector with a single length 1x0 run, and a single length 2x0 run; and single length 3x0 run; and so on:

$vec = ''; 
$n=-1; 
vec( $vec, $n+=$_, 1 ) = 1 for 2 .. 17; 
print unpack 'b*', $vec;;
0100100010000100000100000010000000100000000100000000010000000000100000
+000000100000000000010000000000000100000000000000100000000000000010000
+0000000000001000...
# 2  5   9    14    20     27      35       44        54         65   
+       77           90            104            119             135 
+             152
[download]

Thus, a search for

• 1 x 0 should be found at position 0;
• 2 x 0 should be found at position 2;
• 3@5; 4@9; 5@14; ...

Running your code looking for these runs against that vector produces:

wrog  1: 0
wrog  2: 14
wrog  3: 5
wrog  4: 14
wrog  5: 14
wrog  6: 20
wrog  7: 27
wrog  8: 35
wrog  9: 44
wrog  10: 54
wrog  11: 65
wrog  12: 77
wrog  13: 90
wrog  14: 90
wrog  15: 90
wrog  16: 90
wrog  17: 90
[download]

I believe this is because little-endian bytes are bytes swapped and word swapped relative to big-endian.

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