模板类里面的模板函数的C ++特化

时间：2011-02-14 16:49:28

标签： c++ templates syntax template-specialization

用于专门化模板类中的模板函数的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;
}

我一直在努力猜测专业化的正确语法至少一个小时，并且无法解决任何可编译的问题。任何帮助将不胜感激！

6 个答案:

答案 0 :(得分：18)

AFAIK（以及标准专家可以纠正我），你不能专门化类模板的模板化功能而不专门化类本身...

即。以下我认为可行：

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

答案 1 :(得分：7)

C ++没有对函数模板进行部分特化的概念。但是，通过函数重载，您可以获得与完全特化相同的效果。

我假设你有这样的东西，这实际上是唯一的方法之一。

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东西，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，并且根本不会使用内部模板。

答案 2 :(得分：6)

所以，我采取了不同的方法来回答你的问题。我将从你想要的东西开始，并且有效。然后，也许我们可以弄清楚如何将它置于更接近你真正想要的东西：

#include <string>
#include <iostream>

int getIntThing(const ::std::string &param);

template <typename returnT>
returnT getThingFree(const ::std::string &param);

template <>
int getThingFree<int>(const ::std::string &param)
{
   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';
}

这是另一种有用的想法，并不是你想要的，但更接近。我想你自己已经想过了。它在使用类型演绎的方式上也相当丑陋。

#include <string>
#include <iostream>

template <class c1> class X;

int getIntThing(const ::std::string &param)
{
   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 &param,
                            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 &param) {
      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 &param) {
      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。

答案 3 :(得分：1)

对于好奇，这可能是我将在我自己的代码中使用的解决方案。这是Omnifarious答案的略微变化，无需额外的课程。我仍然为Omnifarious提供道具，因为他完成了大部分的腿部工作：

#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;
}

答案 4 :(得分：0)

尝试

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

这仍然无法编译，但它比你更接近。

我认为你不能集体专注于成员。在开始专门化其成员之前，您必须指定类模板的特定专业化。

答案 5 :(得分：-1)

这是迄今为止我见过的最简单，最简单的方法：

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
}
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