Re: container traits

From:
"Greg Herlihy" <greghe@pacbell.net>
Newsgroups:
comp.std.c++
Date:
Sat, 2 Sep 2006 14:56:57 CST
Message-ID:
<1157224916.695634.59610@i3g2000cwc.googlegroups.com>
Jens Theisen wrote:

Greg Herlihy wrote:

T must be a complete type when std::list<T> is instantiated.


Meaning the following is indeed undefined, as the OP suspected.

 >> struct T;
 >>
 >> typedef std::list<T>::iterator tlist_iterator; // <- UB


Except for the fact that the typedef does not instantiate
std::list<T>::iterator. The type argument for std::list needs to be
complete only at the point of its instantiation, if there is one. (the
latter is a requirement of the Standard Library and not of class
templates in general).

At the point of tlist_iterator's typedef declaration, the compiler has
no idea whether tlist_iterator is ever used subsequently - and
therefore has no idea whether instantiating std::list<T>::iterator is
necessary to complete the program. And unless the instantiation in
question is necessary, the compiler is not allowed to instantiate the
template class implicitly.

Furthermore, the T class can be just as easily implemented without a
typedef:

    #include <list>

    struct T
    {
         std::list<T>::iterator mIter;
    };

    int main()
    {
          T t; // std::list<T>::iterator instantiated here

          ...
    }

With or without the typedef, the program needs T's type to be complete
no earlier than the first line of main - the point at which a T object
is first declared. And clearly, T is a complete type at that point.

After all, if a class could not declare a member function with its own
type as a parameter, the following declaration would be undefined:

    #include <tr1/memory>

    struct A
    {
        std::tr1::shared_ptr<A> next;
    };

In other words an class object would not be able to hold a shared
pointer to its own type.

Greg

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