Question

我正在尝试编写一组通用数学实用程序类（根查找器，集成器等），这些类在构造时接收指向基类的指针，该基类定义了我希望特定算法运行的函数。基类应该只定义可以由用户根据需要实现的公共虚拟接口（抽象或具有默认的简单功能）type operator()(type inputArg)。这将允许用户仅实现所需的仿函数并执行所需的计算。我的母亲在下面：

第一个头定义了抽象接口类

// BaseFunctor.h

#ifndef _BASE_FUNCTOR_H_
#define _BASE_FUNCTOR_H_

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

这是其中一个解算器方法的类

// 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

这是解算器实现： // 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;
}

我做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;
}

我试着让它尽可能简短，如果太长，我很抱歉。基本上我得到错误：

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

如何解决此问题，保留所需功能或为各种所需功能派生类？

由于

Answer 1

rhs和rhsPrime是指针。您需要引用它们才能调用函数调用操作符。

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

如果需要rhs和rhsPrime（即不能为NULL）并且在NewtonsMethod对象具有构造函数后无法更改，则应将它们声明为引用而不是指针。这也消除了取消引用它们来调用函数调用操作符的需要。

下面的示例显示了如何使用引用来引用仿函数。

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...

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
}

C ++定义虚拟基类重载运算符

1 个答案: