Until now the keyword tt(typename) has been used to indicate a template type
parameter. However, it is also used to
    hi(typename: disambiguating code)
 disambiguate code inside templates. Consider the following function template:
        verb(    template <typename Type>
    Type function(Type t)
    {
        Type::Ambiguous *ptr;

        return t + *ptr;
    })

When this code is processed by the compiler, it complains with an -at
first sight puzzling- error message like:
        verb(    4: error: 'ptr' was not declared in this scope)

The error message is puzzling as it was the programmer's intention to
declare a pointer to a type tt(Ambiguous) defined within the class template
tt(Type). But the compiler, confronted with tt(Type::Ambiguous) may interpret
the statement in various ways. Clearly it cannot inspect tt(Type) itself
trying to uncover tt(Type)'s true nature as tt(Type) is a template
type. Because of this tt(Type)'s actual definition isn't available yet.

    The compiler is confronted with two possibilities: either
tt(Type::Ambiguous) is a em(static member) of the as yet mysterious template
tt(Type), or it is a em(subtype) of tt(Type). As the standard
specifies that the compiler must assume the former, the statement
        verb(    Type::Ambiguous *ptr;)

is interpreted as a em(multiplication) of the static member
tt(Type::Ambiguous) and the (now undeclared) entity tt(ptr). The reason for
the error message should now be clear: in this context tt(ptr) is unknown.

    To disambiguate code in which an identifier refers to a
        hi(template: identifying subtypes)
        hi(class template: identifying subtypes)
        hi(typename: and template subtypes)
 subtype of a template type parameter the keyword tt(typename) must be
used. Accordingly, the above code is altered into:
        verb(    template <typename Type>
    Type function(Type t)
    {
        typename Type::Ambiguous *ptr;

        return t + *ptr;
    })

Classes fairly often define subtypes. When such subtypes appear inside
template definitions as subtypes of template type parameters the tt(typename)
keyword em(must) be used to identify them as subtypes. Example: a class
template tt(Handler) defines a tt(typename Container) as its template type
parameter. It also defines a data member storing the iterator returned by the
container's tt(begin) member. In addition tt(Handler) offers a constructor
accepting any container supporting a tt(begin) member.  tt(Handler)'s class
interface could then look like this:
        verb(    template <typename Container>
    class Handler
    {
        Container::const_iterator d_it;

        public:
            Handler(Container const &container)
            :
                d_it(container.begin())
            {}
    };)

What did  we have in mind when designing this class?
    itemization(
    it() The typename tt(Container) represents any container supporting
iterators.
    it() The container presumably supports a member tt(begin). The
initialization tt(d_it(container.begin())) clearly depends on the
template's type parameter, so it's only checked for basic syntactic
correctness.
    it() Likewise, the container presumably supports a em(subtype)
tt(const_iterator), defined in the class tt(Container).
    )
    The final consideration is an indication that tt(typename) is required. If
this is omitted and a tt(Handler) is instantiated the compiler produces a
peculiar compilation error:
        verb(    #include "handler.h"
    #include <vector>
    using namespace std;

    int main()
    {
        vector<int> vi;
        Handler<vector<int> > ph{ vi };
    }
    /*
        Reported error:

    handler.h:4: error: syntax error before `;' token
    */)

Clearly the line
        verb(    Container::const_iterator d_it;)

in the class tt(Handler) causes a problem. It is interpreted by the
compiler as a em(static member) instead of a subtype. The problem is
solved  using tt(typename):
        verb(    template <typename Container>
    class Handler
    {
        typename Container::const_iterator d_it;
        ...
    };)

An interesting illustration that the compiler indeed assumes tt(X::a) to
be a member tt(a) of the class tt(X) is provided by the error message we get
when we try to compile tt(main) using the following implementation of
tt(Handler)'s constructor:
        verb(    Handler(Container const &container)
    :
        d_it(container.begin())
    {
        size_t x = Container::ios_end;
    }
    /*
        Reported error:

        error: `ios_end' is not a member of type `std::vector<int,
                std::allocator<int> >'
    */)

Now consider what happens if the function template introduced at the
beginning of this section doesn't return a tt(Type) value, but a
tt(Type::Ambiguous) value. Again, a subtype of a template type is referred to,
and tt(typename) must be used:
        verb(    template <typename Type>
    typename Type::Ambiguous function(Type t)
    {
        return t.ambiguous();
    })

    tt(Typename)s can be embedded in using declarations. As is often the case,
this
        hi(template: embedded in using declarations)
reduces the complexities of declarations and definitions appearing
elsewhere. In the next example the type tt(Iterator) is defined as a subtype
of the template type tt(Container). tt(Iterator) may now be used without
requiring the use of the keyword tt(typename):
        verb(    template <typename Container>
    class Handler
    {
        using Iterator = Container::const_iterator;

        Iterator d_it;
        ...
    };)

