Re: VC8 compiler behavior?

The code below does not compile with VC8. The error is:

sample.cpp(36) : error C2679: binary '<<' : no operator found which takes
a right-hand operand of type 'const std::vector<_Ty>' (or there is no
acceptable conversion)
        with
        [
            _Ty=int
        ]

but it is clearly there. Is this conformant behavior or a bug?

================= begin code =====================
// sample.cpp

#include <vector>
#include <iostream>

// move the operator below inside the namespace, then it will compile ...
template <class T>
std::ostream& operator<<(std::ostream& ostr, const std::vector<T>& v)
{
  // output comma delimited vector elements ...
  std::vector<T>::const_iterator end = v.end();
  for (std::vector<T>::const_iterator it = v.begin(); it != end; ++it) {
    ostr << *it;
    if (it + 1 != end) ostr << ", ";
  }
  return ostr;
}

namespace formatter {

  template <class T>
  class bracketed {
    public:
      bracketed(const T& t) : value(t) {}
      const T& value;

    private:
      // not implemented ...
      bracketed<T>& operator=(const bracketed<T>&);
  };

  template <class T>
  std::ostream& operator<<(std::ostream& ostr, const bracketed<T>& v)
  {
    // enclose value in brackets ...
    return ostr << '[' << v.value << ']';
  }
};

int main()
{
  using formatter::bracketed;

  // easy example ...
  int x = 21014;
  std::cout << bracketed<int>(x) << std::endl;

  // more complicated example ...
  std::vector<int> v;
  v.push_back(2);
  v.push_back(1);
  v.push_back(0);
  v.push_back(1);
  v.push_back(4);

  std::cout << bracketed<std::vector<int> >(v) << std::endl;
}