+// Split the surname into prefix ("von-part") and family name
+pair<docstring, docstring> parseSurname(docstring const & sname)
+{
+ // Split the surname into its tokens
+ vector<docstring> pieces = getVectorFromString(sname, from_ascii(" "));
+ if (pieces.size() < 2)
+ return make_pair(docstring(), sname);
+
+ // Now we look for pieces that begin with a lower case letter.
+ // All except for the very last token constitute the "von-part".
+ docstring prefix;
+ vector<docstring>::const_iterator it = pieces.begin();
+ vector<docstring>::const_iterator const en = pieces.end();
+ bool first = true;
+ for (; it != en; ++it) {
+ if ((*it).empty())
+ continue;
+ // If this is the last piece, then what we now have is
+ // the family name, notwithstanding the casing.
+ if (it + 1 == en)
+ break;
+ char_type const c = (*it)[0];
+ // If the piece starts with a upper case char, we assume
+ // this is part of the surname.
+ if (!isLower(c))
+ break;
+ // Nothing of the former, so add this piece to the prename
+ if (!first)
+ prefix += " ";
+ else
+ first = false;
+ prefix += *it;
+ }
+
+ // Reconstruct the family name.
+ // Note that if we left the loop with because it + 1 == en,
+ // then this will still do the right thing, i.e., make surname
+ // just be the last piece.
+ docstring surname;
+ first = true;
+ for (; it != en; ++it) {
+ if (!first)
+ surname += " ";
+ else
+ first = false;
+ surname += *it;
+ }
+ return make_pair(prefix, surname);
+}
+
+