C++模板类中模板函数的特化

Question

用于专门化模板类内的模板函数的 C++ 语法是什么？例如，考虑我有以下两个类及其用法。我希望能够为不同类型提供方法 X::getAThing() 的专门实现。例如：int、std::string、任意指​​针或类等。

template <class c1> class X {
public:
   template<typename returnT> returnT getAThing(std::string param);
   static std::string getName();
private:
   c1 theData;
};

// This works ok...
template <class c1> std::string X<c1>::getName() {
   return c1::getName();
}

// This blows up with the error:
// error: prototype for 'int X<c1>::getAThing(std::string)' does not match any in class 'X<c1>'
template <class c1> template <typename returnT> int X<c1>::getAThing(std::string param) {
   return getIntThing(param); // Some function that crunches on param and returns an int.
}

// More specialized definitions of getAThing() for other types/classes go here...

class Y {
public:
   static std::string getName() { return "Y"; }
};

int main(int argc, char* argv[])
{
   X<Y> tester;
   int anIntThing = tester.getAThing<int>(std::string("param"));
   cout << "Name: " <<  tester.getName() << endl;
   cout << "An int thing: " << anIntThing << endl;
}

我已经尝试猜测专业化的正确语法至少一个小时了，但无法弄清楚任何可以编译的内容。任何帮助将不胜感激！

Answer 1

AFAIK（标准专家可以纠正我），如果不专门化类本身，您就无法专门化类模板的模板化函数......

即我认为以下内容可行：

template <> template <> int X<Y>::getAThing<int>(std::string param) {
   return getIntThing(param); // Some function that crunches on param and returns an int.
}

Answer 2

C++ 没有函数模板的部分特化的概念。但是，您可以通过函数重载获得与完全专业化相同的效果。

我假设你有这样的东西，这确实是唯一的方法之一。

template<class TYPE>
class MyInterface {
public:
    template<class RETURN>
    RETURN myFunction(RETURN& ref, ....);
};

在这种情况下，您可以通过声明具有所需类型的普通成员函数来专门化“myFunction()”。 C++的函数重载规则应该给你你想要的，例如

template<class TYPE>
class MyInterface {
public:
    template<class RETURN>
    RETURN myFunction(RETURN& ref, ....);

    // String specialization
    std::string myFunction(std::string& ref, ...);
};

编译器将在适当的地方使用“std::string”函数，并且可能根本不会使用内部模板。

Answer 3

所以，我采取不同的方法来回答你的问题。我将从某种能满足您的要求并且有效的东西开始。然后也许我们可以弄清楚如何将其排列成更接近您真正想要的东西：

#include <string>
#include <iostream>

int getIntThing(const ::std::string ¶m);

template <typename returnT>
returnT getThingFree(const ::std::string ¶m);

template <>
int getThingFree<int>(const ::std::string ¶m)
{
   return getIntThing(param);
}

// More specialized definitions of getAThing() for other types/classes
// go here...

template <class c1> class X {
public:
   template<typename returnT> returnT getAThing(std::string param);
   static std::string getName();
private:
   c1 theData;
};

// This works ok...
template <class c1> std::string X<c1>::getName() {
   return c1::getName();
}

// This also works, but it would be nice if I could explicitly specialize
// this instead of having to explicitly specialize getThingFree.
template <class c1>
template <class RT>
RT X<c1>::getAThing(std::string param) {
   // Some function that crunches on param and returns an RT.
   // Gosh, wouldn't it be nice if I didn't have to redirect through
   // this free function?
   return getThingFree<RT>(param);
}

class Y {
public:
   static std::string getName() { return "Y"; }
};

int main(int argc, char* argv[])
{
   using ::std::cout;
   X<Y> tester;
   int anIntThing = tester.getAThing<int>(std::string("param"));
   cout << "Name: " <<  tester.getName() << '\n';
   cout << "An int thing: " << anIntThing << '\n';
}

这是另一个可行的想法，虽然不完全是您想要的，但更接近。我想你自己已经想到了。它使用类型推导的方式也相当丑陋。

#include <string>
#include <iostream>

template <class c1> class X;

int getIntThing(const ::std::string ¶m)
{
   return param.size();
}

// You can partially specialize this, but only for the class, or the
// class and return type. You cannot partially specialize this for
// just the return type. OTOH, specializations will be able to access
// private or protected members of X<c1> as this class is declared a
// friend.
template <class c1>
class friendlyGetThing {
 public:
   template <typename return_t>
   static return_t getThing(X<c1> &xthis, const ::std::string ¶m,
                            return_t *);
};

// This can be partially specialized on either class, return type, or
// both, but it cannot be declared a friend, so will have no access to
// private or protected members of X<c1>.
template <class c1, typename return_t>
class getThingFunctor {
 public:
   typedef return_t r_t;

   return_t operator()(X<c1> &xthis, const ::std::string ¶m) {
      return_t *fred = 0;
      return friendlyGetThing<c1>::getThing(xthis, param, fred);
   }
};

template <class c1> class X {
public:
   friend class friendlyGetThing<c1>;

   template<typename returnT> returnT getAThing(std::string param) {
      return getThingFunctor<c1, returnT>()(*this, param);
   }
   static std::string getName();
private:
   c1 theData;
};

// This works ok...
template <class c1> std::string X<c1>::getName() {
   return c1::getName();
}

class Y {
public:
   static std::string getName() { return "Y"; }
};

template <class c1>
class getThingFunctor<c1, int> {
 public:
   int operator()(X<c1> &xthis, const ::std::string ¶m) {
      return getIntThing(param);
   }
};

// More specialized definitions of getAThingFunctor for other types/classes
// go here...

int main(int argc, char* argv[])
{
   using ::std::cout;
   X<Y> tester;
   int anIntThing = tester.getAThing<int>(std::string("param"));
   cout << "Name: " <<  tester.getName() << '\n';
   cout << "An int thing: " << anIntThing << '\n';
}

我建议在半私有实用程序命名空间中声明 getThingFunctor 和 FriendlyGetThing。

Answer 4

对于好奇的人来说，这可能是我将在自己的代码中采用的解决方案。这与 Omnifarious 的答案略有不同，无需额外的课程。我仍然支持 Omnifarious，因为他做了大部分的跑腿工作：

#include <iostream>
#include <string>

using namespace std;

// IMPORTANT NOTE: AdaptingFunctor has no access to the guts of class X!
// Thus, this solution is somewhat limited.
template<typename t1> class AdaptingFunctor {
public:
   t1 operator() (string param);
};

// Can specialize AdaptingFunctor for each type required:
template<> int AdaptingFunctor<int>::operator() (string param)
{
   return param.size(); // <=== Insert required type-specific logic here
}

// Additional specializations for each return type can go
// here, without requiring specialization of class c1 for X...


template <class c1> class X {
public:
   template<typename returnT>  returnT getAThing(std::string param)
      {
     AdaptingFunctor<returnT> adapter;
     return adapter(param);
      }
   static std::string getName();
private:
   c1 theData;
};

// Template definition of class method works ok...
template <class c1> std::string X<c1>::getName() {
   return c1::getName();
}

class Y {
public:
   static std::string getName() { return "Y"; }
};


int main(int argc, char* argv[])
{
   X<Y> tester;
   int anIntThing = tester.getAThing<int>(std::string("param"));
   cout << "Name: " <<  tester.getName() << endl;
   cout << "An int thing: " << anIntThing << endl;
}

Answer 5

尝试

template <>
template <class T>
int X<T>::template getAThing<int>(std::string param)
{
   return getIntThing(param);
}

这个仍然无法编译，但它比你的更接近。

我认为你不能让全体成员专业化。您必须先指定类模板的特定特化，然后才能开始特化其成员。

Answer 6

这是迄今为止我见过的最简单、最容易的方法：

template <class T1>
struct MyClass {
  template <class T2>
  void MyFunction();
};

template <class T1>
template <class T2>
void MyClass<T1>::MyFunction() {  // NOTE:  NO T2 on this line.
  // Code goes here
}