C++模板部分特化 - 仅特化一个成员函数

Question

遇到另一个模板问题：

问题：我想部分专门化容器类（foo）来处理对象是指针的情况，并且我只想专门化删除方法。应该如下所示：

lib 代码

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..."); }
};

用户代码

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

myptrs.addSome (new myclass());

这会导致编译器告诉我 myptrs 没有名为 addSome 的方法。 为什么 ？

谢谢。

---

解决方案

based on tony's answer here the fully compilable stuff

---

库

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..."); }
};

用户

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

第二种解决方案（正确的一个）

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..."); } };
};

根据 Core，该解决方案是有效的问题#727。

---

第一个（不正确的）解决方案：（将其保留为评论所引用的内容）

您不能仅专门化课程的一部分。在您的情况下，最好的方法是重载函数 deleteSome ，如下所示：

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

另一个解决方案。使用辅助函数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 3

我还没有看到这个解决方案，使用boost的 启用_if，is_same 和 remove_pointer 获取类中的两个函数，没有任何继承或其他缺陷。

请参阅下面的仅使用 remove_pointer 的版本。

#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..."); 
    }
};

一个简化的版本是：

#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..."); 
    }
};

它适用于 MSVC 9：（注释掉给出错误的行，因为它们不正确，但很适合测试）

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);
}

Answer 4

为单个函数创建基类 deleteSome

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

进行部分特化

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

使用您的基类

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

Answer 5

您可以使用继承来使其发挥作用：

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..."); }
};