[LyX/master] Fix problem noted at bug #10582.

[Richard Heck]

commit 322331c53aa4f9aecc7a4e170b7cafee987a823b
Author: Richard Heck <rgh...@lyx.org>
Date:   Sun Mar 12 16:43:15 2017 -0400

    Fix problem noted at bug #10582.
    
    When we have a name with more than two parts, but no "von",
    it was coming out as, e.g.:
        Obama, Barack Hussain Obama
    i.e., with the last name appearing twice.
    
    Also adds a check for names without spaces, which would have given:
        Pele, Pele
    
    This was not the original issue at #10582, so that bug is still
    open (though I cannot reproduce it).
---
 src/BiblioInfo.cpp |   21 +++++++++++++++------
 1 files changed, 15 insertions(+), 6 deletions(-)

diff --git a/src/BiblioInfo.cpp b/src/BiblioInfo.cpp
index 6437ccf..0f76628 100644
--- a/src/BiblioInfo.cpp
+++ b/src/BiblioInfo.cpp
@@ -65,15 +65,18 @@ pair<docstring, docstring> nameParts(docstring const & name)
        // include the "von" part. This isn't perfect.
        // Split on spaces, to get various tokens.
        pieces = getVectorFromString(name, from_ascii(" "));
-       // If we only get two, assume the last one is the last name
-       if (pieces.size() <= 2)
+       // unusual not to have a space, but could happen
+       if (pieces.size() < 2)
+               return make_pair(from_ascii(""), name);
+       // If we get two, assume the last one is the last name
+       if (pieces.size() == 2)
                return make_pair(pieces.front(), pieces.back());
 
        // Now we look for the first token that begins with
        // a lower case letter or an opening group {.
        docstring prename;
        vector<docstring>::const_iterator it = pieces.begin();
-       vector<docstring>::const_iterator en = pieces.end();
+       vector<docstring>::const_iterator const en = pieces.end();
        bool first = true;
        for (; it != en; ++it) {
                if ((*it).empty())
@@ -81,6 +84,12 @@ pair<docstring, docstring> nameParts(docstring const & name)
                char_type const c = (*it)[0];
                if (isLower(c) || c == '{')
                        break;
+               // if this is the last time through the loop, then
+               // what we now have is the last name, so we do not want
+               // to add that to the prename.
+               if (it + 1 == en)
+                       break;
+               // add this piece to the prename
                if (!first)
                        prename += " ";
                else
@@ -88,10 +97,10 @@ pair<docstring, docstring> nameParts(docstring const & name)
                prename += *it;
        }
 
-       if (it == en) // we never found a "von" or group
-               return make_pair(prename, pieces.back());
-
        // 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) {