C / C ++ - 牛顿方法的迭代计数器

时间:2012-11-17 17:03:53

标签: c++ c function printf

我创建了一个运行牛顿方法的函数,用于近似函数的解(定义为f)。我的函数返回更好的根的近似值,但是它不会正确显示函数中执行的迭代次数。

这是我的代码:

#include <stdio.h> 
#include <math.h> 
#include <cstdlib>
#include <iostream>

double newton(double x_0, double newtonaccuracy);

double f(double x);

double f_prime(double x);

int main() 
{
   double x_0;  

   double newtonaccuracy;  

   int converged;  

   int iter;

   printf("Enter the initial estimate for x : ");

   scanf("%lf", &x_0);

   _flushall();

   printf("\n\nEnter the accuracy required : ");

   scanf("%lf", &newtonaccuracy);

   _flushall();


   if (converged == 1) 
      {
        printf("\n\nNewton's Method required %d iterations for accuracy to %lf.\n", iter, newtonaccuracy);

        printf("\n\nThe root using Newton's Method is x = %.16lf\n", newton(x_0, newtonaccuracy));
      } 

   else 
      {
        printf("Newton algorithm didn't converge after %d steps.\n", iter);
      }



      system("PAUSE");
} 


double newton(double x_0, double newtonaccuracy) 
{
   double x = x_0;

   double x_prev;

   int iter = 0;


   do 
   {
      iter++;


      x_prev = x;

      x = x_prev - f(x_prev)/f_prime(x_prev);

   } 
   while (fabs(x - x_prev) > newtonaccuracy && iter < 100);

   if (fabs(x - x_prev) <= newtonaccuracy)
   {
      int converged = 1;
   }  
   else
   {
      int converged = 0; 
   }   




    return x;
}  


double f(double x) {
       return ( cos(2*x) - x );
}  

double f_prime(double x) 
{
   return ( -2*sin(2*x)-1 ); 
}  

尽可能具体,这就是行:

printf("\n\nNewton's Method required %d iterations for accuracy to %lf.\n", iter, newtonaccuracy);
那是给我带来麻烦的。每次我运行这个程序时都会说“Newton's Method需要2686764次迭代......”但是如果我编码正确(我的代码允许的最大迭代次数为100),这不可能是真的。

1 个答案:

答案 0 :(得分:2)

iter中使用的变量main未在newton函数中初始化或使用,您使用局部变量iter。您需要通过引用将iter传递给newton,或者找到一种方法将其从函数中返回。

以下是通过引用获取某些参数并对其进行修改的函数示例:

double foo(double& initial_value, int& iterations)
{
  initial_value *= 3.14159;
  iterations = 42;
  return initial_value/2.;
}

来自来电方:

double x + 12345.;
int iter = 0;
double y = foo(initial_value, iter);