2014/1/9 Sebastian Pfützner <s.pfuetzner@onlinehome.de>
Hello,
the following code does not compile with the last line in main(). The error message is
main.cpp: In function 'int main()': main.cpp:37:8: error: 'boost::detail::operator_brackets_result<It, S, S&>::type' has no member named 'i' it[3].i = 0; ^
What could be the problem?
#include <boost/iterator/iterator_facade.hpp>
struct S { int i; float f; };
class It : public boost::iterator_facade<It, S, boost::random_access_ traversal_tag> { private: friend class boost::iterator_core_access;
static S s;
reference dereference() const { return s; }
bool equal(It const&) const { return true; }
void increment() {} void decrement() {} void advance(difference_type n) {}
difference_type distance_to(It const&) const { return 0; } };
int main() { It it; (*(it + 3)).i = 0; S& s = it[3]; s.i = 0; it[3].i = 0; }
Hello, In your case, it[3] returns a boost::detail::operator_brackets_proxy<It> instead of S& you expect: // A proxy return type for operator[], needed to deal with // iterators that may invalidate referents upon destruction. // Consider the temporary iterator in *(a + n) template <class Iterator> class operator_brackets_proxy; As the comment sais, it[3] won't return a reference directly "to deal with // iterators that may invalidate referents upon destruction. // Consider the temporary iterator in *(a + n)". I don't know in what situations that can happen. I don't see a clean way to force it[3] to return a real reference. Can anyone comment on this? That said, here's a hack for you: right after the definition of struct S insert this code: class It; namespace boost { namespace detail { template <> struct operator_brackets_result<It,S,S&> { typedef S& type; }; }} (Your code won't link as it is, buth that's irrelevent here) HTH Kris