2

我正在尝试使用二分法求解一个二次方程。尝试评估根时出现此错误：“没有匹配的调用函数”。

#include "assign4.h"
#include <iostream>

using namespace std;

int main(int argc, char * argv[]){
   solution s;
   double root;

   cout << "Enter interval endpoints: ";
   cin >> s.xLeft >> s.xRight;

   cout << "Enter tolerance: ";
   cin >> s.epsilon;

   root = s.bisect (s.xLeft, s.xRight, s.epsilon, s.f, s.error);

   if (!(s.error))
      cout << "Root found at " << root << "\nValue of f(x) at root is: " << s.f(root);
   else
      cout << "The solution of a quadratic equation with coefficients: " << endl;
      cout << "a = " << a << ", b = " << b << ", c = " << c << endl;
      cout << "has not been found." << endl;

   return 0;
}

错误发生在 root = ... 我的函数 f 似乎有问题但我不明白出了什么问题。下面两段代码是我的类和类实现文件。我们刚刚开始使用类，所以我不确定我的问题是在那里还是只是在上面的代码中。

#ifndef ASSIGN4_H
#define ASSIGN4_H

class solution {

public:
   double xLeft, xRight;
   double epsilon;
   bool error;

   double bisect(double, double, double, double f(double), bool&);
   double f(double);
};
#endif // ASSIGN4_H

///////////////////////////////////////// ///////////////////////////////////////// ///////////////////////////////////////// ////////////

#include "assign4.h"
#include <iostream>
#include <cmath>

using namespace std;

double solution::bisect (double xLeft, double xRight, double epsilon, double func(double), bool& error) {
   double xMid;
   double fLeft, fRight;
   double fMid;

   fLeft = f(xLeft);
   fRight = f(xRight);

   error = (fLeft * fRight) > 0;
   if (error)
      return -999.0;

   while (fabs (xLeft - xRight) > epsilon) {
      xMid = (xLeft + xRight) / 2.0;
      fMid = f (xMid);

      if (fMid == 0.0)
         return xMid;
      else if (fLeft * fMid < 0.0)
         xRight = xMid;
      else
         xLeft = xMid;

      cout << "New Interval is [" << xLeft << ", " << xRight << "]" << endl;
   }

return (xLeft + xRight) / 2.0;
}

double solution::f (double x) {
   return ((5 * pow(x,2.0)) + (5 * x) + 3);
}
4

4 回答 4

3

第4个参数是函数指针，

double bisect(double, double, double, double f(double), bool&);

当你调用这个函数时：

root = s.bisect (s.xLeft, s.xRight, s.epsilon, s.f, s.error);

虽然成员虚构double f(double)与该参数的类型不同，因为这是 C++ 成员函数而不是静态的，因此在编译时添加了“this”参数这个成员函数。

type 将静态关键字添加到函数中。

于 2014-04-15T20:14:42.057 回答
2

函数指针的语法通常是：double (*f)(double). 除此之外，您还试图通过非成员函数指针传递成员函数。由于您的函数不使用任何成员变量，因此最简单的解决方案是static：

class solution {
  // ...
  static double f(double);
};
于 2014-04-15T20:14:43.283 回答
1

如果要使用指向成员函数的指针。

改变

double bisect(double, double, double, double f(double), bool&);

至

double bisect(double, double, double, double (solution::*f)(double), bool&);

在声明和定义中。

将呼叫从

root = s.bisect (s.xLeft, s.xRight, s.epsilon, s.f, s.error);

至

root = s.bisect (s.xLeft, s.xRight, s.epsilon, &solution::f, s.error);

这就是我为我成功编译和链接的东西。

 #include <iostream>
 #include <typeinfo>
 #include <math.h>

 using namespace std;

 class solution {

 public:
    double xLeft, xRight;
    double epsilon;
    bool error;

    double bisect(double, double, double, double (solution::*f)(double), bool&);
    double f(double);
 };

 using namespace std;

 double solution::bisect (double xLeft, double xRight, double epsilon, double (solution::*func)(double), bool& error) {
    double xMid;
    double fLeft, fRight;
    double fMid;

    fLeft = (this->*func)(xLeft);
    fRight = (this->*func)(xRight);

    error = (fLeft * fRight) > 0;
    if (error)
       return -999.0;

    while (fabs (xLeft - xRight) > epsilon) {
       xMid = (xLeft + xRight) / 2.0;
       fMid = (this->*func)(xMid);

       if (fMid == 0.0)
          return xMid;
       else if (fLeft * fMid < 0.0)
       {
          xRight = xMid;
          fRight = fMid;
       }
       else
       {
          xLeft = xMid;
          fLeft = fMid;
       }

       cout << "New Interval is [" << xLeft << ", " << xRight << "]" << endl;
    }

 return (xLeft + xRight) / 2.0;
 }

 double solution::f (double x) {
    return ((5 * pow(x,2.0)) + (5 * x) + 3);
 }

 int main(int argc, char * argv[]){
    solution s;
    double root;

    cout << "Enter interval endpoints: ";
    cin >> s.xLeft >> s.xRight;

    cout << "Enter tolerance: ";
    cin >> s.epsilon;

    root = s.bisect (s.xLeft, s.xRight, s.epsilon, &solution::f, s.error);

    if (!(s.error))
       cout << "Root found at " << root << "\nValue of f(x) at root is: " << s.f(root) << endl;
    else
    {
       cout << "The solution of a quadratic equation with coefficients: " << endl;
       // cout << "a = " << a << ", b = " << b << ", c = " << c << endl;
       cout << "has not been found." << endl;
    }
    return 0;
 }
于 2014-04-15T20:24:37.123 回答
1

我相信这与您的回调函数有关。通常，当您使用不正确的函数调用时，您会遇到这种编译器错误。如果您想要这种回调函数，您可能需要查看函数指针。

http://www.cprogramming.com/tutorial/function-pointers.html

于 2014-04-15T22:30:00.147 回答