Hi, This *is* the correct behaviour. There were several discussions on this topic in the past (you can search archive for more insight). split always returns n+1 tokens where n is a number of separators. This is deterministic and allows to reconstruct the original sequence. Best Regards, Pavol luca regini wrote:

The following code:

using namespace boost::algorithm; std::string test("DataItem.item[4]"); std::vector< std::string > path; split(path,test,is_any_of(".[]"),token_compress_on); for(int i=0;i<path.size();i++) { const char* dbg=path[i].c_str(); std::cout<<path[i]<<std::endl; } std::cout<<" DONE!"<<std::endl;

produces the following output:

DataItem 4 <null string> DONE!

This happens every time the test string ends with a terminator. Is this the desired behaviour? It seems to me that it is more intuitive if the split returns just the first two items without the final <null> string.

Regards, Luca Regini _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost