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daniel.w.gelder@gmail.com
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 Posted: Wed Apr 27, 2005 11:45 pm Post subject: Recursive partial specialization |
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Hi. I have a template with 3 params:
template <typename BASE, typename SUPER=void, typename SUPER2=void>
Coolness
{
};
typedef Coolness<long> LongCoolness;
typedef Coolness<long, bool> LongAndBoolCoolness;
OK, this is the trick here. When the template is typedefed in this
form:
typedef Coolness<BASE, Coolness Thing;
IE with a Coolness as a second parameter, I then want T1 and T2 to be
the parameters instead:
typedef Coolness<BASE, T1, T2> Thing;
Can I accomplish with partial specialization?
Thanks very much
Dan
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Kanenas
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| Posted: Mon May 30, 2005 1:44 am Post subject: Re: Recursive partial specialization |
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On 27 Apr 2005 16:45:58 -0700, [email]daniel.w.gelder (AT) gmail (DOT) com[/email] wrote:
| Quote: | Hi. I have a template with 3 params:
template <typename BASE, typename SUPER=void, typename SUPER2=void
|
Don't forget a 'struct' or 'class' keyword.
| Quote: | Coolness
{
};
typedef Coolness
typedef Coolness<long, bool> LongAndBoolCoolness;
OK, this is the trick here. When the template is typedefed in this
form:
typedef Coolness<BASE, Coolness Thing;
IE with a Coolness as a second parameter, I then want T1 and T2 to be
the parameters instead:
typedef Coolness<BASE, T1, T2> Thing;
Can I accomplish with partial specialization?
You can't get type equivalence, but you can close if you also use |
inheritance.
template <BASE, typename T1, typename T2>
class Coolness<BASE, Coolness
: public Coolness<BASE, T1, T2>
{
typedef Coolness<T1, T2> Super1;
typedef Coolness<BASE, T1, T2> Base;
/* These constructors may or may not make any sense
for Coolness.
*/
Coolness(const BASE& b, const T1& t1, const T2& t2)
: Base(b, t1, t2) {}
Coolness(const BASE& b, const Super1& s)
};
Any template instantiations which have a 1 or 2 arg Coolness as the
second parameter, e.g. 'Coolness<string, Coolness
string> >', will use the specialization. Note that the inner
Coolness-es in the example ('Coolness<string>' and
'Coolness<Coolness') do not use the specialization.
However, 'Coolness<string, Coolness >' has
2 Coolness-es which do use the specialization.
Kanenas
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