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。
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);