Find a type in a parameter pack

You can hide this implementation in a namespace and use another class that calls your implementation with a default parameter example:

namespace detail
{
    // your code as it is in the question
}

template <typename A, typename... B>
struct index_of
{
    static int const value = detail::index_of<0, A, B...>::value;
};

Edit

In his comment DyP suggests a simpler way to default I using an alias

template <typename A, typename... B>
using index_of = detail::index_of<0, A, B...>;

template <typename A, typename B, typename... C>
struct index_of
{
  static constexpr int const value =
    std::is_same<A, B>{}
    ? 0
    : (index_of<A, C...>::value >= 0) ? 1+index_of<A, C...>::value : -1;
};

template <typename A, typename B>
struct index_of<A, B>
{
  static constexpr int const value = std::is_same<A, B>{} -1;
};

Note the std::is_same<A, B>{} -1 uses a conversion from bool to int.

---

Better by deriving from integral_constant:

template <typename A, typename B, typename... C>
struct index_of
  : std::integral_constant
    < int,
        std::is_same<A, B>{}
      ? 0
      : (index_of<A, C...>{} == -1 ? -1 : 1+index_of<A, C...>{})
    >
{};

template <typename A, typename B>
struct index_of<A, B>
  : std::integral_constant < int, std::is_same<A, B>{} -1 >
{};

---

If you don't need to return -1 in case the type isn't found: (if anyone knows how to incorporate a static_assert here for a pretty diagnostic message, I'd appreciate a comment/edit)

template <typename A, typename B, typename... C>
struct index_of
  : std::integral_constant < std::size_t,
                             std::is_same<A, B>{} ? 0 : 1+index_of<A, C...>{} >
{};

template <typename A, typename B>
struct index_of<A, B>
  : std::integral_constant<std::size_t, 0>
{
    constexpr operator std::size_t() const
    {
        return   std::is_same<A, B>{}
               ? 0
               : throw std::invalid_argument("Type not found!");
    }
};