some observations, mainly about the N>=8 case and if you're going to do this with regexps:
  1. regexps are designed for searching forward; while there are backtracking/"look-behind" operators, they're rather ad hoc and sometimes a bad idea to use,

  2. the byte-classes you need for for "ends with a sequence of exactly n zero bits", that, because of (1), you'll want to use for anchoring your regexps, are much easier to construct if your bytes are taken to be little-endian (i.e., the high order bits of byte k are taken to be adjacent to the low order bits of byte k+1).
    E.g., the range for "ends with a sequence of exactly 1 zero bit" is 0x40-0x7f, "...2 zero bits" is 0x20-0x3f and so on.
    And while this messes up the corresponding expressions/classes for "begins with at least n zero bits", the "at least" vs. "exactly" makes a difference in that you now only have to check the zeroness of the first (N - n) mod 8 bits, so I think things are still easier in the little-endian world.
Putting this all together we need a |-join of 8 regexps (n = 0..7 and m = N - n mod 8) of the form 
[2^(7-n) .. (2^(7-n+1)-1)] \0{N/8} (?: ($ch mod 2^m == 0) | "\0+" )

[download]
modulo this not being quite correct perl-regexp code but you know how to fix that.

In reply to Re: Locating a specified number of contiguous 0 bits within a large bitstring efficiently. by wrog
in thread Locating a specified number of contiguous 0 bits within a large bitstring efficiently. by BrowserUk

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