C++ template partial specialization - specializing one member function only

Question

Hi everybody,

Bumped into another templates problem:

The problem: I want to partially specialize a container-class (foo) for the case that the objects are pointers, and i want to specialize only the delete-method. Should look like this:

The lib code

template <typename T>
class foo
{
public:
    void addSome    (T o) { printf ("adding that object..."); }
    void deleteSome (T o) { printf ("deleting that object..."); }
};

template <typename T>
class foo <T *>
{
public:
    void deleteSome (T* o) { printf ("deleting that PTR to an object..."); }
};

The user code

foo<myclass> myclasses;
foo<myclass*> myptrs;

myptrs.addSome (new myclass());

This results into the compiler telling me that myptrs doesnt have a method called addSome. Why ?

Thanx.

---

Solution

based on tony's answer here the fully compilable stuff

---

lib

template <typename T>
class foobase
{
public:
    void addSome    (T o) { printf ("adding that object..."); }
    void deleteSome (T o) { printf ("deleting that object..."); }
};


template <typename T>
class foo : public foobase<T>
{ };

template <typename T>
class foo<T *> : public foobase<T *>
{
public:
    void deleteSome (T* o) { printf ("deleting that ptr to an object..."); }
};

user

foo<int>    fi;
foo<int*>   fpi;

int   i = 13;

fi.addSome (12);      
fpi.addSome (&i);

fpi.deleteSome (12);     // compiler-error: doesnt work
fi.deleteSome (&i);   // compiler-error: doesnt work
fi.deleteSome (12);   // foobase::deleteSome called
fpi.deleteSome (&i);     // foo<T*>::deleteSome called

Answer 1

Second solution (correct one)

template <typename T>
class foo
{
public:
    void addSome    (T o) { printf ("adding that object..."); } 
    void deleteSome(T o) { deleteSomeHelper<T>()(o); }
protected:
    template<typename TX> 
    struct deleteSomeHelper { void operator()(TX& o) { printf ("deleting that object..."); } };
    template<typename TX> 
    struct deleteSomeHelper<TX*> { void operator()(TX*& o) { printf ("deleting that PTR to an object..."); } };
};

This solution is valid according to Core Issue #727.

---

First (incorrect) solution: (kept this as comments refer to it)

You cannot specialize only part of class. In your case the best way is to overload function deleteSome as follows:

template <typename T>
class foo
{
public:
    void addSome    (T o) { printf ("adding that object..."); }
    void deleteSome (T o) { printf ("deleting that object..."); }
    void deleteSome (T* o) { printf ("deleting that object..."); }
};

Answer 2

You can use inheritance to get this to work :

template <typename T>
class foobase
{
public:
    void addSome    (T o) { printf ("adding that object..."); }
    void deleteSome (T o) { printf ("deleting that object..."); }
};

template <typename T>
class foo : public foobase<T>
{ };

template <typename T>
class foo <T *> : public foobase<T>
{
public:
    void deleteSome (T* o) { printf ("deleting that PTR to an object..."); }
};

Answer 3

Create base class for single function deleteSome

template<class T>
class base {
public:
  void deleteSome (T o) { printf ("deleting that object..."); }
}

Make partial specialization

template<class T>
class base<T*> {
public:
  void deleteSome (T * o) { printf ("deleting that PTR to an object..."); }
}

Use your base class

template <typename T>
class foo : public base<T> {
 public:    
   void addSome    (T o) { printf ("adding that object..."); 
}

Answer 4

Another solution. Use the auxiliary function deleteSomeHelp.

template <typename T>
class foo {
 public:    
   void addSome    (T o) { printf ("adding that object..."); 
   template<class R>
   void deleteSomeHelp (R   o) { printf ("deleting that object..."); }};
   template<class R>
   void deleteSomeHelp (R * o) { printf ("deleting that PTR to an object..."); }};
   void deleteSome (T o) { deleteSomeHelp(o); }
}

Answer 5

I haven't seen this solution yet, using boost's enable_if, is_same and remove_pointer to get two functions in a class, without any inheritance or other cruft.

See below for a version using only remove_pointer.

#include <boost\utility\enable_if.hpp>
#include <boost\type_traits\is_same.hpp>
#include <boost\type_traits\remove_pointer.hpp>

template <typename T>
class foo
{
public:
    typedef typename boost::remove_pointer<T>::type T_noptr;

    void addSome    (T o) { printf ("adding that object..."); }

    template<typename U>
    void deleteSome (U o, typename boost::enable_if<boost::is_same<T_noptr, U>>::type* dummy = 0) { 
        printf ("deleting that object..."); 
    }
    template<typename U>
    void deleteSome (U* o, typename boost::enable_if<boost::is_same<T_noptr, U>>::type* dummy = 0) { 
        printf ("deleting that PTR to that object..."); 
    }
};

A simplified version is:

#include <cstdio>
#include <boost\type_traits\remove_pointer.hpp>

template <typename T>
class foo
{
public:
    typedef typename boost::remove_pointer<T>::type T_value;

    void addSome    (T o) { printf ("adding that object..."); }

    void deleteSome (T_value& o) { // need ref to avoid auto-conv of double->int
        printf ("deleting that object..."); 
    }

    void deleteSome (T_value* o) { 
        printf ("deleting that PTR to that object..."); 
    }
};

And it works on MSVC 9: (commented out lines that give errors, as they are incorrect, but good to have for testing)

void main()
{
   foo<int> x;
   foo<int*> y;

   int a;
   float b;

   x.deleteSome(a);
   x.deleteSome(&a);
   //x.deleteSome(b); // doesn't compile, as it shouldn't
   //x.deleteSome(&b);
   y.deleteSome(a);
   y.deleteSome(&a);
   //y.deleteSome(b);
   //y.deleteSome(&b);
}