问题
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 stufflib
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
回答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..."); }
};
回答2:
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); }
}
回答3:
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);
}
回答4:
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...");
}
回答5:
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..."); }
};
来源:https://stackoverflow.com/questions/1757791/c-template-partial-specialization-specializing-one-member-function-only