Can you achieve fn(x1 ^ fn(x0)) with a fold expression?

Question

Is it possible to achieve the following using a fold expression?

template<class... Args>
auto foo(Args... args)
{
    //calling foo(x0, x1, x2) should be exactly equivalent to
    //calling fn(x2 ^ fn(x1 ^ fn(x0)))
}

Answer 1

If you insist on a fold expression, something along these lines could probably be made to work (not tested):

template <typename T>
struct Wrapper {
  T& val;
};

template <typename T, typename U>
auto operator^(Wrapper<T> l, Wrapper<U> r) {
  return Wrapper(r.val ^ fn(l.val));
}

template<class... Args>
auto foo(Args... args)
{
  return fn((... ^ Wrapper<Args>(args)).val);
}