2014/1/9 Sebastian Pfützner

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::type' has no member named 'i' it[3].i = 0; ^

What could be the problem?

#include

struct S { int i; float f; };

class It : public boost::iterator_facade { 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 { typedef S& type; }; }} (Your code won't link as it is, buth that's irrelevent here) HTH Kris