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考虑以下来自 GNU 网站的关于minimzation的示例。假设fn1我需要从一组点中插入目标而不是给定目标，(xa,ya)以获得我的目标函数double fn2(double x){gsl_interp_eval( myinterp, xa[],ya[],x)。问题是如何避免每次调用时都必须myinterp在目标函数内设置插值对象，而是将其作为参数传递给目标。fn2我试过一个结构，但没有运气。

我想一个更一般的问法是说“如何将类型的对象传递给类型gsl_interp的函数gsl_function？”

#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;
}
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1 回答 1

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对不起，我犯了一个错误，希望不会浪费任何人的时间。我在结构中声明了类似

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

即我的 gsl_interp 对象的名称是int. 这不是很聪明，因为int当然是类型声明。

于 2013-07-24T12:35:15.830 回答