通过指向虚拟基类的指针访问模板类的模板成员函数

Accessing template member functions of template class through pointer to virtual base class?

本文关键字：成员 函数 访问 虚拟 基类 指针      更新时间：2023-10-16

我正在尝试实现一系列模板化对象，这些对象将通过非模板虚拟基类指针进行访问。简化后，基类看起来像：

class Base
{
public:
  virtual void printThing(const int &thing) = 0;
  virtual void printThing(const double &thing) = 0;
  virtual void printThing(const bool &thing) = 0;
};

我想做的事情在下面的派生类实现中进行了概述：

#include <iostream>
template <typename T>
class Derived : public Base
{
public:
  void printThing(const T &thing);
  template <typename U>
  void printThing(const U &thing);
};
template <typename T>
void Derived<T>::printThing(const T &thing)
{
  std::cout << "Derived same type " << thing << std::endl;
}
template <typename T>
template <typename U>
void Derived<T>::printThing(const U &thing)
{
  std::cout << "Derived diff type " << thing << std::endl;
}
template <>
template <>
void Derived<double>::printThing(const int &thing)
{
  std::cout << "Derived<double> specialized for int " << thing << std::endl;
}

这对任何类型的U都有效——只要代码直接在Derived的实例上调用成员函数，并且U在编译时是已知的。

但是，当我试图通过指向Base的指针访问Derived时，我会遇到编译器错误，如下面的测试程序所示：

int main(int argc, char* argv[])
{
  Derived<int> dint;
  Derived<double> ddouble;
  Base * bint = &dint;
  Base * bdouble = &ddouble;
  double d = 3.14;
  int i = 42;
  bint->printThing(i);
  bint->printThing(d);
  bdouble->printThing(i);
  bdouble->printThing(d);
  return 0;
}

clang++在Mac OS 10.8.5上给出了这样的反馈：

razor:playpen cfry$ clang++ template-specialization.cc 
template-specialization.cc:43:16: error: variable type 'Derived<int>' is an abstract class
  Derived<int> dint;
               ^
template-specialization.cc:7:16: note: unimplemented pure virtual method 'printThing' in 'Derived'
  virtual void printThing(const double &thing) = 0;
               ^
template-specialization.cc:8:16: note: unimplemented pure virtual method 'printThing' in 'Derived'
  virtual void printThing(const bool &thing) = 0;
               ^
template-specialization.cc:44:19: error: variable type 'Derived<double>' is an abstract class
  Derived<double> ddouble;
                  ^
template-specialization.cc:6:16: note: unimplemented pure virtual method 'printThing' in 'Derived'
  virtual void printThing(const int &thing) = 0;
               ^
template-specialization.cc:8:16: note: unimplemented pure virtual method 'printThing' in 'Derived'
  virtual void printThing(const bool &thing) = 0;
               ^
2 errors generated.
razor:playpen cfry$

请注意，我已经显式地实现了Derived<double>::printThing(const int &)，编译器声称它不存在并且广义的Derived<T>::printThing(const U &)成员函数似乎没有被实例化。

有没有任何可移植的方法可以告诉编译器，我打算为每个"未实现"的虚拟方法实例化通用模板成员函数？

我已经尝试了很多替代方案，但到目前为止，唯一有效的方案是为基类成员函数提供一个默认实现，并编写Derived的包装器，该包装器为所需的U类型显式实现printThing((。

我使用Eli Bendersky在这里解释的Curioly Recurring Template Pattern(CRTP(找到了答案。

它需要添加另一层模板类：

template <class Child>
class Adapter : public Base
{
public:
  void printThing(const int &thing)
  {
    static_cast<Child *>(this)->printThingInternal(thing);
  }
  void printThing(const double &thing)
  {
    static_cast<Child *>(this)->printThingInternal(thing);
  }
  void printThing(const bool &thing)
  {
    static_cast<Child *>(this)->printThingInternal(thing);
  }
};
template <typename T>
class Derived : public Adapter<Derived <T> >
{
public:
  void printThingInternal(const T &thing);
  template <typename U>
  void printThingInternal(const U &thing);
};

有了这个添加，以及从Adapter继承Derived的简单更改，该程序安抚了编译器，更好的是，生成了我想要的结果：

razor:playpen cfry$ clang++ template-specialization.cc 
razor:playpen cfry$ ./a.out
Derived same type 42
Derived diff type 3.14
Derived<double> specialized for int 42
Derived same type 3.14
razor:playpen cfry$
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