[Boost-users] [boost::transform_iterator] failure with std::vector::const_iterator

From: Ovanes Markarian (om_boost_at_[hidden])
Date: 2007-03-04 14:32:02

Hello!

 

I have an assertion failure using std::vector::const_iterator and
boost::transform_iterator.

I have a constant seq of constant string pairs delimited by special
character (here ':'). My application needs only once to iterate through
these pairs and return iterator ranges from the beginning until the colon
and after colon until the end. I thought a transform iterator can do the
job. I get in the marked position an assertion failure using MSVC 2005
Express Edition.

Here the code snippet with problem description:

#include <string>
#include <vector>
#include <utility>
#include <functional>

#include <boost/iterator/transform_iterator.hpp>
#include <boost/range/iterator_range.hpp> 
#include <boost/utility.hpp> 
#include <boost/bind.hpp>

typedef boost::iterator_range<std::string::const_iterator>
iter_range;
typedef std::pair<iter_range,iter_range>
result_type;

struct string_splitter : std::unary_function<std::string const&,
result_type> 
{
        result_type operator()(std::string const& s)const throw()
        {
                iter_range r1(s.begin(), s.begin() + s.find(":"));
                iter_range r2(r1.end()+1, s.end());
                return result_type(r1,r2);
        }
};

typedef std::vector<std::string>
str_vector;
typedef str_vector::const_iterator
vector_citer;
typedef boost::transform_iterator<string_splitter, vector_citer>
msg_opcode_citer;

int main()
{
        //create vector of strings and put 2 strings inside
        str_vector v;
        v.push_back(std::string("test1:opcode1"));
        v.push_back(std::string("test2:opcode2"));

        //!!!increment operator or dereference operator cause an assertion
failure!!!
        // stating: "vector iterators incompatible"
        for(msg_opcode_citer curr = msg_opcode_citer(const_cast<str_vector
const&>(v).begin(), string_splitter()), end; 
                        curr != end; ++curr)
        {
                result_type t = *curr;
        }

        return 0;
}

Did I miss something?

 

With Kind Regards,

Ovanes Markarian