Re: Wierd template class

[..]
Ok the void operator()(){} now contains this code:

  void operator()() {
    typedef typename std::vector<E>::iterator R;
    R answer = NAMESPACE::find(a.begin(), a.end(), s);
  }

But R is not defined as a template parameter in the template class
definition:

template <typename E>
class Mytest {
    public:
    Mytest(int n) {
        s = E(0);
        a.resize(n,E(0));

    }

  void operator()() {
    typedef typename std::vector<E>::iterator R;
    R answer = ::find(a.begin(), a.end(), s);
  }

    private:
          E s;
          std::vector<E> a;

};

In this context I assume R is just a name for the iterator, but how
can an iterator have 2 names: "R answer = ...."?