Re: syntax error in std::vector<T>::const_iterator where T = (class) U<V>

From:

"Bo Persson" <bop@gmb.dk>

Newsgroups:

comp.lang.c++

Date:

Mon, 7 Feb 2011 18:49:07 +0100

Message-ID:

<8rapk3FskcU1@mid.individual.net>

Juha Nieminen wrote:

Bo Persson <bop@gmb.dk> wrote:

That the compiler cannot formally know that
std::vector<A<U>>::const_iterator is a type (because someone could
specialize std::vector for some user defined types). You have to
add a 'typename' to specify what it is:

Is there any conrete example where the same dependent name could be
used as both a type name and a non-type name, and this could
potentially cause confusion? Why exactly is the 'typename' keyword
necessary?

Yes, if you specialize a class there is no requirements that the base
template and the specialization should have anything in common (except
their names :-).

template<class T>
class vec
{
public:
    class const_iterator
    {
    public:
       // some members here
    };
};

template<>
class vec<int>
{
public:
   double const_iterator; // silly, but possible
};

Now, what is vec<T>::const_iterator, a type or a member variable?
Depends on what T is!

Bo Persson