使用gsl_interp的gsl_min_fminimizer

Question

考虑来自minimzation上的GNU网站的以下示例。假设我需要从一组点(xa,ya)中插值目标函数来获得目标函数double fn2(double x){gsl_interp_eval( myinterp, xa[],ya[],x)，而不是给定的目标函数。问题是如何避免每次调用时都必须在目标函数fn2内设置插值myinterp对象，而是将其作为参数传递给目标函数。我已经尝试了一个结构，但是没有成功。

我猜更一般的提问方式是“如何将类型为gsl_interp的对象传递给类型为gsl_function的函数？”

#include <stdio.h>
#include <gsl/gsl_errno.h>
#include <gsl/gsl_math.h>
#include <gsl/gsl_min.h>

double fn1 (double x, void * params)
{
  return cos(x) + 1.0;
}

int
main (void)
{
  int status;
  int iter = 0, max_iter = 100;
  const gsl_min_fminimizer_type *T;
  gsl_min_fminimizer *s;
  double m = 2.0, m_expected = M_PI;
  double a = 0.0, b = 6.0;
  gsl_function F;

  F.function = &fn1;
  F.params = 0;

  T = gsl_min_fminimizer_brent;
  s = gsl_min_fminimizer_alloc (T);
  gsl_min_fminimizer_set (s, &F, m, a, b);

  printf ("using %s method\n",
          gsl_min_fminimizer_name (s));

  printf ("%5s [%9s, %9s] %9s %10s %9s\n",
          "iter", "lower", "upper", "min",
          "err", "err(est)");

  printf ("%5d [%.7f, %.7f] %.7f %+.7f %.7f\n",
          iter, a, b,
          m, m - m_expected, b - a);

  do
    {
      iter++;
      status = gsl_min_fminimizer_iterate (s);

      m = gsl_min_fminimizer_x_minimum (s);
      a = gsl_min_fminimizer_x_lower (s);
      b = gsl_min_fminimizer_x_upper (s);

      status 
        = gsl_min_test_interval (a, b, 0.001, 0.0);

      if (status == GSL_SUCCESS)
        printf ("Converged:\n");

      printf ("%5d [%.7f, %.7f] "
              "%.7f %+.7f %.7f\n",
              iter, a, b,
              m, m - m_expected, b - a);
    }
  while (status == GSL_CONTINUE && iter < max_iter);

  gsl_min_fminimizer_free (s);

  return status;
}

Answer 1

对不起，我犯了一个错误，希望没有浪费任何人的时间。我在结构中声明了类似于

struct myint = { double a; gsl_interp int;};

也就是说，我的gsl_interp对象的名字是int。这并不是很聪明，因为int当然是一个类型声明。