Jeremiah Willcock wrote:
On Mon, 12 Dec 2011, Larry Evans wrote:
On 12/12/11 17:04, Larry Evans wrote:
On 12/12/11 16:39, Jeremiah Willcock wrote: [snip] Looking at your code, it seems you solved the problem by removing the nested specialization of is_end, with a specialization in file scope of check_unless_end. I've a vague memory of hearing some similar solution elsewhere. Is there some reason why specializations cannot occur within the class?
-regards, Larry
The reason is explained in this thread:
http://groups.google.com/group/comp.lang.c++.moderated/browse_frm/thread/c04...
I don't see anything in that thread that talks about why member templates can't be specialized in-class. I don't see a good reason for that limitation; how is a class member specialization different from one in a namespace other than that there can be template parameters on the class itself?
-- Jeremiah Willcock
First of all I want to thank everyone who contributed to this thread. I carefully read every single message and the links. I was really lost. I was really bothered with the idea that the template specialization would work outside the class but not in it - as you were. The links didn't seem on point on this subject. I took my original solution and moved the specializations outside the class. This failed because the I wasn't defined so I added it do the template arguments. Then a few syntax corrections and it started to work - OK, problem solved - but now I was left with this really annoying feature whose behavior changed when the specialization was moved out side the class. So I moved it back in - and voila - worked as expected. Here is my solution: template<typename I> struct ForwardIterator { template<typename I, typename B> struct is_end {}; template<typename I2> struct is_end<I2, false_> { typedef typename deref<I2>::type t1; typedef typename next<I2>::type t2; BOOST_CONCEPT_ASSERT(( ForwardIterator<t2> )); }; typedef typename is_end< I, typename boost::is_same< boost::mpl::l_iter<boost::mpl::l_end>, I > > type; typedef typename boost::mpl::print< I >::type t1; BOOST_MPL_ASSERT(( is_convertible< typename I::category, forward_iterator_tag > )); }; Basically, I just had to avoid depending on the "outter" template parameter. This had bothered me before but I wasn't aware of the right syntax to avoid it. I had to use the specialization syntax above rather than the template<> syntax I was familiar with. So I guess that resolves the problem. Again, thanks for everyone who contributed - really would never have found it by myself. Robert Ramey