This is the "leading common substring" problem. There's been work in Perl done on it. Here's a relatively fast solution:

# leading common substring
# returns length of LCS
# requires Perl 5.6+
sub lcs {
  ($_[0] ^ $_[1]) =~ /\0*/;
  return $+[0];
}
[download]

If you're not fortunate enough to have Perl 5.6 yet, here's a compatible approach:

# leading common substring
# returns length of LCS
# requires Perl 5.6+
sub lcs {
  ($_[0] ^ $_[1]) =~ /(\0*)/;
  return length $1;
}
[download]

Update: in case you were worried, I do not need a ^ anchor in these regexes, because * will gladly match zero times at the beginning of the string if it has to.

_____________________________________________________
Jeff[japhy]Pinyan: Perl, regex, and perl hacker.
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

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