Enumerations hi(enumeration: nested) hi(nested enumerations) may also be
nested in classes. Nesting enumerations is a good way to show the close
connection between the enumeration and its class. Nested enumerations have the
same controlled visibility as other class members. They may be defined in the
private, protected or public sections of classes and are inherited by derived
classes.  In the class ti(ios) we've seen values like tt(ios::beg) and
tt(ios::cur). In the current i(GNU) bf(C++) implementation these values are
defined as values of the ti(seek_dir) enumeration:
        verb(    class ios: public _ios_fields
    {
        public:
            enum seek_dir
            {
                beg,
                cur,
                end
            };
    };)

As an illustration assume that a class tt(DataStructure) represents a data
structure that
may be traversed in a forward or backward direction. Such a class can define
an enumeration tt(Traversal) having the values tt(FORWARD) and
tt(BACKWARD). Furthermore, a member function tt(setTraversal) can be defined
requiring a tt(Traversal) type of argument. The class can be defined as
follows:
        verb(    class DataStructure
    {
        public:
            enum Traversal
            {
                FORWARD,
                BACKWARD
            };
            setTraversal(Traversal mode);
        private:
            Traversal
                d_mode;
    };)

Within the class tt(DataStructure) the values of the tt(Traversal)
enumeration can be used directly. For example:
        verb(    void DataStructure::setTraversal(Traversal mode)
    {
        d_mode = mode;
        switch (d_mode)
        {
            FORWARD:
                // ... do something
            break;

            BACKWARD:
                // ... do something else
            break;
        }
    })

Outside of the class tt(DataStructure) the name of the enumeration type is
not used to refer to the values of the enumeration. Here the classname is
sufficient. Only if a variable of the enumeration type is required the name of
the enumeration type is needed, as illustrated by the following piece of code:
        verb(    void fun()
    {
        DataStructure::Traversal                // enum typename required
            localMode = DataStructure::FORWARD; // enum typename not required

        DataStructure ds;
                                                // enum typename not required
        ds.setTraversal(DataStructure::BACKWARD);
    })

In the above example the constant tt(DataStructure;:FORWARD) was used to
specify a value of an enum defined in the class tt(DataStructure). Instead of
tt(DataStructure::FORWARD) the construction tt(ds.FORWARD) is also
accepted. In my opinion this syntactic liberty is ugly: tt(FORWARD) is a
symbolic value that is defined at the class level; it's not a member of tt(ds),
which is suggested by the use of the member selector operator.

    Only if tt(DataStructure) defines a nested class tt(Nested), in
turn defining the enumeration tt(Traversal), the two class scopes are
required. In that case the latter example should have been coded as follows:
        verb(    void fun()
    {
        DataStructure::Nested::Traversal
            localMode = DataStructure::Nested::FORWARD;

        DataStructure ds;

        ds.setTraversal(DataStructure::Nested::BACKWARD);
    })

Here a construction like tt(DataStructure::Nested::Traversal localMode =
ds.Nested::FORWARD) could also have been used, although I personally would
avoid it, as tt(FORWARD) is not a member of tt(ds) but rather a symbol that is
defined in tt(DataStructure).
