Re: Template name as template parameter (or something like that)
On 2008-11-17 10:19:32 -0500, Jon
<nomorespamplease@nospam-3492hdasuds8d8wsd.net> said:
Normally I can search and find answers to things like this but with this
one I'm not even sure what to search for and haven't had any luck. Anyway,
I'm trying to use a template name as a template parameter and can't seem
to figure out what I need to do. Code follows:
template <class MAPTYPE, class KEYTYPE, class VALUETYPE> class double_map
{
MAPTYPE<KEYTYPE,VALUETYPE> forward_; // for "forward" lookups
MAPTYPE<VALUETYPE,KEYTYPE> reverse_; // for "reverse" lookups
...
So I want to use MAPTYPE (which is a template itself)
No, MAPTYPE is the name of a type. That's what the definition of
double_map says. Later in the definition it's used as if it were a
template, which, given the previous definition, is an error.
What you want is for that first parameter to be a template and not a
type. So define the template that way:
template <template<class, class> class maptype, class keytype, class
valuetype> ...
By the way, hash_map almost certainly won't work here, because it
probably takes more then two template arguments, with all but the first
two having defaults. Welcome to the wonderful world of template
template parameters.
Incidentally, a more verbose approach is used in the standard library
(whose interface pre-dated template template parameters). Use your
original definition, and instantiate it with the right type:
double_map<hash_map<string, int>, string, int> mymap;
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)