Re: work round for std::distance's long arm.

From:

Louis Lavery <Louis@laver.demon.co.uk>

Newsgroups:

comp.lang.c++.moderated

Date:

19 Aug 2006 10:32:03 -0400

Message-ID:

<ec6tr2$n3q$1$830fa795@news.demon.co.uk>

Greg Herlihy wrote:

Louis Lavery wrote:

Without the nested typedefs the following fails to compile
under vc7 and gcc 3.2.

/* distance.cpp */
#include <iterator>
#include <list>

namespace user
{
    template<typename Iter,typename Dist> struct Cursor
    {
#if 0
        typedef void iterator_category;
        typedef void difference_type;
        typedef void value_type;
        typedef void reference;
        typedef void pointer;
#endif
    };

    template<typename Iter,typename Dist>
        Dist distance(
        Cursor<Iter,Dist> const&,
        Cursor<Iter,Dist> const&)
    {
        return 0;
    }
}

int main()
{
    user::Cursor<std::list<int>::iterator,int> cur;

    distance(cur,cur); // XXX

    return 0;
}
/* distance.cpp end */

Because Cursor's instantiated with an iterator from std,
I guess what happens (at XXX) is the compiler looks in
namespace std and sees something like...

template<class Iter>
    typename iterator_traits<Iter>::difference_type
    distance<Iter,Iter) {...}

...and so needs to instantiate iterator_traits<Cursor> to get
the return type. But the default iterator_traits requires its
parameter to have nested typedefs for iterator_category etc.
Have I got that right?

The distance() function call does indeed call std::distance().

Actually it doesn't call any thing as the code doesn't compile,
as I say in the first line of my post.

When the void typedefs are exposed, and it does compile, it calls
user:distance() as expected.

However,

iterator_traits is not instantiated in order to find std::distance's
return type (the compiler can figure that out without instantiating
iterator_traits).

Yes it is (no it can't).

It has to instantiate iterator_traits, with Iter == Cursor, in
order to check that what is actually being returned is convertible
to difference_type (which is the function's return type).

Louis.

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