我正在尝试在用于费米气体的c实现工具中建立一个共享库.我正在使用GSL库以数字方式求解函数,并且我的代码运行时没有问题,而没有作为脚本运行,但是尝试将其转换为共享库和类时遇到了问题.
我是c编程的新手,似乎无法适应我的问题.可能因为我不太了解答案.
我的代码是：
/* Define structure for the GSL-function: chempot_integrand */
struct chempot_integrand_params { double mu; double T; };
double
ChemicalPotential::chempot_integrand (double x, void * params){
/* Computes the integrand for the integral used to obtain the chemical potential.
*
* This is a GSL-function, which are integrated using gsl_integration_qag.
*/
// Get input parameters.
struct chempot_integrand_params * p = (struct chempot_integrand_params *) params;
double mu = p->mu;
double T = p->T;
// Initiate output parameters for GSL-function.
gsl_sf_result_e10 result;
int status = gsl_sf_exp_e10_e( ( gsl_pow_2(x) - mu ) / T , &result );
if (status != GSL_SUCCESS){
printf ("Fault in calculating exponential function.");
}
// Return (double) integrand.
return (gsl_pow_2(x) / ( 1 + result.val * gsl_sf_pow_int(10,result.e10) ));
}
/* Define structure for the GSL-function: chempot_integration */
struct chempot_integral_params { double T; };
double
ChemicalPotential::chempot_integration (double mu, double T){
/* Computes the integral used to obtain the chemical potential using the integrand: chempot_integrand.
*/
// Set input parameters for the integrand: chempot_integrand.
struct chempot_integrand_params params_integrand = { mu, T };
// Initiate the numerical integration.
gsl_integration_workspace * w = gsl_integration_workspace_alloc (1000); // Allocate memory for the numerical integration. Can be made larger if neccessary, REMEMBER to change it in the function call: gsl_integration_qag as well.
double result, error;
gsl_function F;
F.function = &ChemicalPotential::chempot_integrand;
F.params = ¶ms_integrand;
// Upper limit for integration
double TOL = 1e-9;
double upp_lim = - T * gsl_sf_log(TOL) + 10;
gsl_integration_qag (&F, 0, upp_lim, 1e-12, 1e-12, 1000, 6, w, &result, &error);
// Free memory used for the integration.
gsl_integration_workspace_free (w);
return result;
}
当编译时出现错误
error: cannot convert ‘double (Fermi_Gas::ChemicalPotential::*)(double, void*)’ to ‘double (*)(double, void*)’
排队
F.function = &ChemicalPotential::chempot_integrand;
解决方法:
人们一遍又一遍地问这个问题确实很有趣.原因之一可能是所提出的解决方案不容易理解.我一个人在理解和实施它们时遇到了问题. (如您所料,解决方案对我来说不是开箱即用的.)
在tlamadon的帮助下,我刚刚找到了一个可能在这里也很有用的解决方案.让我们看看你们的想法.
综上所述,问题在于您有一个包含成员函数的类,您要在该函数上使用GSL库中的内容进行操作.如果GSL界面需要一个
gsl_function F;
所以这是示例类：
class MyClass {
private:
gsl_f_pars *p; // not necessary to have as member
public:
double obj(double x, void * pars); // objective fun
double GetSolution( void );
void setPars( gsl_f_pars * xp ) { p = xp; };
double getC( void ) ; // helper fun
};
该练习的目的是为了能够
>启动MyClass测试,
>为其提供参数结构(或编写相应的构造函数),然后
>在其上调用test.GetSolution(),它将返回用于GSL函数的任何内容(obj,根,整数或任何最小值)
现在的诀窍是在struct gsl_f_pars参数中放置一个元素,该元素是指向MyClass的指针.这是结构：
struct gsl_f_pars {
double a;
double b;
double c;
MyClass * pt_MyClass;
};
最后一步是提供一个包装器,该包装器将在MyClass :: GetSolution()内部调用(包装器是成员函数MyClass :: obj的代表,我们不能仅在类内部使用& obj指向该包装器).该包装器将采用参数struct,取消引用pt_MyClass并评估pt_MyClass的成员obj：
// Wrapper that points to member function
// Trick: MyClass is an element of the gsl_f_pars struct
// so we can tease the value of the objective function out
// of there.
double gslClassWrapper(double x, void * pp) {
gsl_f_pars *p = (gsl_f_pars *)pp;
return p->pt_MyClass->obj(x,p);
}
完整示例太长了,无法在此处发布,因此我提出了要点.它是header file和cpp file,无论您使用的是GSL还是应该都能工作.编译并运行
g++ MyClass.cpp -lgsl -o test
./test