Re: template specialization for pointer-to-type

From:
"Eric Pruneau" <eric.pruneau@cgocable.ca>
Newsgroups:
comp.lang.c++
Date:
Mon, 30 Jun 2008 23:20:52 -0400
Message-ID:
<hohak.13416$Mc.11930@read1.cgocable.net>
".rhavin grobert" <clqrq@yahoo.de> a ?crit dans le message de news:
e272f287-de27-4d6f-bff1-d1f0813b69d0@k30g2000hse.googlegroups.com...

guess you have the following:

_________________________________________________
template <class T>
class CQVector
{
public:
// find an element, returns index or -1 if none is found
int find(int id) const;
private:
std::vector<T> m_vec;
};

template <class T>
int CQVector<T>::find(int id) const
{
int iCnt = m_vec.size();
while (iCnt-->0)
{
if (m_vec[iCnt].ID() == id)
break;
}
return iCnt;
}
_________________________________________________

this finds an element in the vector by calling elements fn ID() and
compiles for all structures/classes that have an ID() memeber-fn
returning something comparable to int.

now i also want the possibility to store pointers, eg

CQVector<MyClass*> m_foo;

and need some specialisation that does a...
_________________________________________________

int iCnt = m_vec.size();
while (iCnt-->0)
{
if (m_vec[iCnt]->ID() == id)
break;
}
return iCnt;
_________________________________________________

what syntax is need ed for the specialisation?
something like template<class*T> doesnt work...

TIA, -.rhavin;)


No need to partially specialize your class by the way. Here an example.

#include <boost/utility.hpp>
#include <boost/type_traits.hpp>

using namespace std;

//This template function will be called only if T is a pointer. return type
is int
template<typename T>
typename boost::enable_if_c<boost::is_pointer<T>::value, int>::type
Do() {return 1;}

//This template function will be called only if T is NOT a pointer. return
type is int
template<typename T>
typename boost::disable_if_c<boost::is_pointer<T>::value, int>::type
Do() {return 2;}

template<typename T>
class A
{
public:
 int DoSomething()
 {
  return Do<T>(); // Do actually take careof doing the right thing
depending on T
 }
};

int main()
{
 A<int> a1;
 A<int*> a2;
 cout <<a1.DoSomething() <<endl; // print 2
 cout <<a2.DoSomething() <<endl; // print 1
return 0;
}

Ok I agree that the Do<T> function template could look a little (ok maybe
not just a little...) strange at first sight but it is not actually that
bad.

take for example

boost::enable_if_c<boost::is_pointer<T>::value, int>::type

there is 2 template arg to enable_if_c
 1.boost::is_pointer<T>::value
 2. int

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