[boost] Re: Partially specializing less<> with nested template class

From: Larry Evans (cppljevans_at_[hidden])
Date: 2004-07-08 16:16:59

On 07/08/2004 03:53 PM, David B. Held wrote:
> Larry Evans wrote:
> 
>> [...]
>> Do you mean like the :0: operator in
>>
>> http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&selm=1072lb8ira3jg41%40corp.supernews.com 
>> [...]
> 
> 
> I didn't totally follow your reasoning, but I think I might have
> a more generic solution that I will post in c.l.c++.m.
> 

Putting it another way, there's a 1 to 1 correspondence between 
templates nested to a depth N, and an N argument template.  IOW,
this is the same as saying there's a 1 to 1 correspondence between
a function curried N times and an N argument function.  Hence,
if you can deduce the template arguments in an N template argument
function, then you can deduce the template arguments in a function
with 1 argument formed from templated nested to a depth N.  IOW,
just as the template arguments in:

template
   < typename A1
   , typename A2
   ...
   , typename AN
   >
void f( T<A1,A2,...,AN>& a)
;

can be deduced; so can the template arguments in:

template
   < typename A1
   , typename A2
   ...
   , typename AN
   >
void f( T1<A1>:0:T2<A2>:0: ... :0:TN<AN>& a)
;

because there's a 1 to 1 correspondence between:

   T<A1,A2,...,AN>

and:

   T1<A1>:0:T2<A2>:0: ... :0:TN<AN>

Since there is such a 1 to 1 correspondence, deducing
1 should be as possible as deducing the other.

At least I think so :)