java - 在 Java 中转换时在 C 代码中出现指针和 if 语句错误

Question

我正在尝试将 C 代码转换为 java，它是 Newton Raphson 算法的实现。一切顺利，但 C 代码中使用的指针存在问题，我已在 Java 中删除了它们。 C代码部分是:

 x = newton(x_0, error, max_iters, &iters, &converged); //Call to a function (newton)

   if (converged) {

printf("Newton algorithm converged after %d steps.\n", iters);

    printf("The approximate solution is %19.16e\n", x);
    printf("f(%19.16e) = %19.16e\n", x, f(x));
   } else 
{

printf("Newton algorithm didn't converge after %d steps.\n", 
          iters);

    printf("The final estimate was %19.16e\n", x);
    printf("f(%19.16e) = %19.16e\n", x, f(x));
  }

函数定义是这样的:

double newton(double x_0, double error, int max_iters, 
          int* iters_p, int* converged_p)

现在的问题是，两个指针变量的值每次都为零。此外，if(converged) 显示了不兼容类型的错误消息。需要 boolean 值，找到 int。下面是 java 代码，请帮助解决这个问题。

 //Member Functions///////
 public
            double function( double x) 
            { 
                return x*x - 2;
            }

            double F_Deriv( double x ) 
            {
                return 2.0*x; 
            }


            double newton(double x_0, double error, int max_iters,int iters, int converged) 
            {
   double x = x_0;
   double x_prev;
   int    iter = 0;

   do {
      iter++;
      x_prev = x;
      x = x_prev - function(x_prev)/F_Deriv(x_prev);
   } 
   while (Math.abs(x - x_prev) > error && iter < max_iters);

   if (Math.abs(x - x_prev) <= error)
      converged = 1;
   else
      converged = 0;
   iters = iter;

   return x;
} 


/////Main Function///////
    public static void main(String[] args) {

        Newton_Raphson obj=new Newton_Raphson();

        Scanner input=new Scanner(System.in);


        double x_0;       /* Initial guess                */
        double x;         /* Approximate solution         */
        double error;       /* Maximum error                */
        int    max_iters; /* Maximum number of iterations */
        int    iters;     /* Actual number of iterations  */
        int    converged; /* Whether iteration converged  */

        System.out.println( "Enter Initial Solution: " );
        x_0=input.nextDouble();

        System.out.println( "Enter Error: " );
        error=input.nextDouble();


        System.out.println( "Enter Maximum Iterations: " );
        max_iters=input.nextInt();

        x = obj.newton(x_0, error, max_iters, iters, converged);

        if(converged) 
        {
            System.out.println("Newton algorithm converged after "+ iters +" steps.");
            System.out.println("The approximate solution is "+ x);

        } 

        else 
        {
            System.out.println("Newton algorithm didn't converge after " + iters + " steps.");
            System.out.println("The final estimate was " + x);

        }

    }

Answer 1

Java 按值传递参数，所以

if (Math.abs(x - x_prev) <= error)
    converged = 1;
else
    converged = 0;
iters = iter;

不会更改调用者传递的参数。这些更改永远不会离开被调用的函数。

模仿输出参数的最简单方法，resp。 C中传递指针，就是传递一个单长数组，

double newton(double x_0, double error, int[] max_iters,int iters, boolean[] converged)

并设置(和查询)iters[0] resp。 收敛[0]。