我需要帮助改进我的C程序代码，以使用runge-kutta四阶方法求解微分方程

Question

我需要c代码的帮助。代码实际上运行fyn，但我需要它做更多的工作。该程序是使用龙格-库塔四阶方法求解微分方程组的代码。长话短说，我所需要的只是一种从界面输入问题的方式，而不是从代码内部输入。请帮忙，我们将不胜感激。

#define X0 1.0
#define Xn 2.0
#define STEPLEN 10
#define Y1 1 //initial condition for y1 at x=0
#define Y2 0 //initial condition for y2 at x=0
#‎include <stdio.h>
#include <math.h>
#include <stdlib.h>
double f1(double x,double y1,double y2)
{
    //a function to solve the functions 2 using x and y
    return(y2);
}
double f2(double x,double y1,double y2)
{
    //a function to solve the functions 1 using x and y
    return(x*log(x) - (2*y1)/(x*x) + (2*y2)/x);
}
main()
{
    double x,y1[STEPLEN],y2[STEPLEN];
    double xmid,k1[2],k2[2],k3[2],k4[2],h = (double)(Xn - X0)/STEPLEN;
    x = X0;
    y1[0]=Y1;
    y2[0]=Y2;
    int i;
    printf("This are the initial values. y1=%.10f y2=%.10f and h=%fnn", y1[0], y2[0],h);
    //printf("xt|k1t|k2t|k3t|k4t|ynn");
    for(i=0; i<STEPLEN; i++){
        k1[0]=h * f1(x,y1[i],y2[i]);
        k1[1]=h * f2(x,y1[i],y2[i]);
        xmid = x + h/2.0;
        k2[0] = h * f1(xmid, y1[i] + k1[0] * 1/2.0, y2[i] + k1[1] * 1/2.0);
        k2[1] = h * f2(xmid, y1[i] + k1[0] * 1/2.0, y2[i] + k1[1] * 1/2.0);
        k3[0] = h * f1(xmid, y1[i] + k2[0] * 1/2.0, y2[i] + k2[1] * 1/2.0);
        k3[1] = h * f2(xmid, y1[i] + k2[0] * 1/2.0, y2[i] + k2[1] * 1/2.0);
        k4[0] = h * f1(x + h, y1[i] + k3[0] , y2[i] + k3[1]);
        k4[1] = h * f2(x + h, y1[i] + k3[0] , y2[i] + k3[1]);
        y1[i+1] = y1[i] + (k1[0] + 2*k2[0] + 2*k3[0] + k4[0])/6;
        y2[i+1] = y2[i] + (k1[1] + 2*k2[1] + 2*k3[1] + k4[1])/6;
        printf("When x=%fn",x+h);
        printf("k1 for y1=%.10fn",k1[0]);
        printf("k1 for y2=%.10fn",k1[1]);
        printf("k2 for y1=%.10fn",k2[0]);
        printf("k2 for y2=%.10fn",k2[1]);
        printf("k3 for y1=%.10fn",k3[0]);
        printf("k3 for y2=%.10fn",k3[1]);
        printf("k4 for y1=%.10fn",k4[0]);
        printf("k4 for y2=%.10fn",k4[1]);
        printf("y1 when x is %f=t%.10fn",x+h,y1[i+1]);
        printf("y2 when x is %f=t%.10fnn",x+h,y2[i+1]);
        // printf("%f %f %f %f %f %f %f %f %f %f n",k1[0],k1[1],k2[0],k2[1],k3[0],k3[1],k4[0],k4[1],y1[i+1],y2[i+1]);
        x = x+h;
    }
    system("pause");
}

Answer 1

可以简单地执行

    double X0, Xn;
    fputs("Enter X0n", stdout);
    if (sscanf("%lf", &XO) != 1) Handle_EOForInputProblem();
    fputs("Enter Xnn", stdout);
    if (sscanf("%lf", &Xn) != 1) Handle_EOForInputProblem();
    ... 

或者创建一个助手函数来提示、处理用户I/O问题，并读入各种double。

int Mayor_Read_Double(const char *prompt, double *dest) {
  char buffer[100];
  for (;;) {
    fputs(prompt, stdout);
    if (fgets(buffer, sizeof buffer, stdin) == NULL) {
      return -1;
    }
    char *endptr;
    errno = 0;
    double number = strtod(buffer, &endptr);
    if (errno && number != 0.0) {
      fputs("Input out of rangen", stdout);
      continue;
      }
    if (buffer == endptr) {  // no conversion
      fputs("Invalid inputn", stdout);
      continue;
      }
    // Detect extra garbage
    while (isspace((unsigned char) *endptr)) endptr++;
    if (*endptr) {
      fputs("Extra inputn", stdout);
      continue;
    }
    *dest = number;
    return 1;
  }
}
double X0, Xn;
if (Mayor_Read_Double("Enter X0n", &X0) != 1) return 0;
if (Mayor_Read_Double("Enter Xnn", &Xn) != 1) return 0;
  ...