How to typedef a type derived through several layers of templates?

Question

Maybe my Google-fu just isn't strong enough.

Using GCC 4.4.3, I've got a set of classes like this:

template <typename storage_t, typename index_t = std::size_t, typename
    leaf_payload_t = std::size_t>
struct btree_node {
    public:
    typedef btree_node<storage_t, index_t, leaf_payload_t> this_t;
    typedef boost::shared_ptr<this_t> handle_t;

    // [...]
};

template <typename storage_t, typename index_t = std::size_t, typename
    leaf_payload_t = std::size_t>
class btree {
    public:
    class caching_storage_t;
    typedef btree_node<caching_storage_t, index_t, leaf_payload_t> node_t;
    typedef typename node_t::handle_t nodehandle_t;

    // [...]

    class caching_storage_t {
        public:
        //typedef typename btree::nodehandle_t nodehandle_t; // Fails -- why?
        typedef typename boost::shared_ptr<node_t> nodehandle_t;

        // [...]
    };
};

As you can see, I've had to redefine nodehandle_t in caching_storage_t, because if I try it with the commented-out typedef line (which I'd prefer), I get an error "no type named ‘handle_t’ in ‘struct btree_node<...>’" -- which is obviously incorrect, and the compiler knows it since the typedef works fine in btree. I've also tried a using typename btree::nodehandle_t;, and every variation I've been able to think of on both, to no avail.

Is this a language/syntax problem (and if so, what is the right syntax), or is it a compiler bug?

(There's a similar question here, but it doesn't seem to apply because the thing I'm trying to typedef isn't itself a template. Nothing else I've been able to find seems even close.)

Answer 1

It looks like a compiler bug to me. To be sure, I put your code into clang and instantiated btree<int>::caching_storage_t, all working fine also with the comment chars removed.

It also works on GCC4.5.1 and GCC4.3.4.