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我正在尝试编写一组通用数学实用程序类（根查找器、积分器等），它们在构造一个指向基类的指针时接受，该基类定义了我希望特定算法对其进行操作的函数。基类应该只定义一个type operator()(type inputArg)可由用户根据需要实现的公共虚拟接口（抽象或具有默认琐碎功能）。这将允许用户只实现所需的函子并执行所需的计算。我的mwe如下：

第一个头文件定义了抽象接口类

// BaseFunctor.h

#ifndef _BASE_FUNCTOR_H_
#define _BASE_FUNCTOR_H_

class BaseFunctor
{
public:
   virtual double operator() (double x) = 0;
};
#endif

这是求解器方法之一的类

// NewtonsMethod.h

#ifndef _NEWTONS_METHOD_H_
#define _NEWTONS_METHOD_H_

class BaseFunctor;

class NewtonsMethod
{
public: 

   NewtonsMethod(BaseFunctor *rhsIn,
                 BaseFunctor *rhsPrimeIn,
                 double       x0In);

   ~NewtonsMethod();

   bool DetermineRoot(double &root);

 private:

   double       x0;
   BaseFunctor *rhs;
   BaseFunctor *rhsPrime;

   static const double       EPSILON;
   static const unsigned int MAX_ITER;

};
#endif

这是求解器实现： // NewtonsMethod.cpp

#include "NewtonsMethod.h"
#include "BaseFunctor.h"
#include <cmath>

const double       NewtonsMethod::EPSILON  = 1e-9;
const unsigned int NewtonsMethod::MAX_ITER = 30;

NewtonsMethod::NewtonsMethod(BaseFunctor *rhsIn,
                             BaseFunctor *rhsPrimeIn,
                             double       x0In) :
   rhs     (rhsIn),
   rhsPrime(rhsPrimeIn),
   x0      (x0In)
{ }

NewtonsMethod::~NewtonsMethod() { }

bool NewtonsMethod::DetermineRoot(double &root)
{
   // This is obviously not implemented
   root = rhs(1.0) / rhsPrime(2.0);
   return false;
}

以及我进行派生类实现的主要功能：// Main.cpp

#include "BaseFunctor.h"
#include "NewtonsMethod.h"
#include <iostream>
#include <iomanip>

class fOfX : public BaseFunctor
{
   double operator() (double x)
   {
      return x * x - 2.0;
   }
};

class fPrimeOfX : public BaseFunctor
{
   double operator() (double x)
   {
      return 2.0 * x;
   }
};


int main()
{
   double x0 = 2.0;

   BaseFunctor *f      = new fOfX();
   BaseFunctor *fPrime = new fPrimeOfX(); 

   NewtonsMethod newton(f, fPrime, x0);

   double root      = 0.0;
   bool   converged = newton.DetermineRoot(root);

   if (converged)
   {
      std::cout << "SUCCESS: root == " << std::setprecision(16) << root << std::endl;
   }
   else
   {
      std::cout << "FAILED: root == " << std::setprecision(16) << root << std::endl;
   }
   delete f;
   delete fPrime;
}

我试图让它尽可能简短，如果太长，请见谅。基本上我得到了错误：

g++ Main.cpp NewtonsMethod.cpp -o main
NewtonsMethod.cpp: In member function ‘bool NewtonsMethod::DetermineRoot(double&)’: 
NewtonsMethod.cpp:29: error: ‘((NewtonsMethod*)this)->NewtonsMethod::rhs’ cannot be used    as a function
NewtonsMethod.cpp:29: error: ‘((NewtonsMethod*)this)->NewtonsMethod::rhsPrime’ cannot be   used as a function

我怎样才能解决这个问题，保持所需的功能或为各种需要的功能派生一个类？

谢谢

4

1 回答 1

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rhs并且rhsPrime是指针。您需要引用它们才能调用函数调用运算符。

(*rhs)(1.0) / (*rhsPrime)(2.0)

如果rhs和rhsPrime是必需的（即不能为NULL）并且在对象具有构造函数后不能更改，NewtonsMethod则应将它们声明为引用而不是指针。这也将消除取消引用它们以调用函数调用运算符的需要。

下面的例子展示了如何使用引用来引用函子。

class NewtonsMethod
{
public: 
   NewtonsMethod(BaseFunctor& rhsIn,
                 BaseFunctor& rhsPrimeIn,
                 double       x0In);

   ~NewtonsMethod();

   bool DetermineRoot(double &root);

 private:

   double       x0;
   BaseFunctor& rhs;
   BaseFunctor& rhsPrime;

   static const double       EPSILON;
   static const unsigned int MAX_ITER;
};

int main()
{
   double x0 = 2.0;

   fOfX       f;
   fPrimeOfX  fPrime;

   NewtonsMethod newton(f, fPrime, x0);
}

...或者...

int main()
{
   double x0 = 2.0;

   BaseFunctor *f      = new fOfX();
   BaseFunctor *fPrime = new fPrimeOfX(); 

   NewtonsMethod newton(*f, *fPrime, x0);

   // ... other code including delete for the functors
}
于 2013-07-04T01:16:38.843 回答