将gsl与c ++一起使用时如何避免静态成员函数

2022-01-03 00:00:00 callback c++ c++11 gsl std-function

我想在 C++ 类中使用 GSL，而不将成员函数声明为 static.这样做的原因是因为我不太了解它们，而且我不确定线程??安全性.从我读到的内容来看，std::function 可能是一个解决方案，但我不确定如何使用它.

I would like to use GSL within a c++ class without declaring member functions as static. The reason for this is because I don't know them too well and I'm not sure about thread safety. From what I read, std::function might be a solution but I'm not sure how to use it.

我的问题归结为如何删除 g 声明中的 static?

My question comes down to how can I remove static in declaration of g?

#include<iostream>
#include <functional>
#include <stdlib.h>
#include <gsl/gsl_math.h>
#include <gsl/gsl_monte.h>
#include <gsl/gsl_monte_plain.h>
#include <gsl/gsl_monte_miser.h>
#include <gsl/gsl_monte_vegas.h>


using namespace std;

class A {
public:
  static double g (double *k, size_t dim, void *params)
  {
    double A = 1.0 / (M_PI * M_PI * M_PI);
    return A / (1.0 - cos (k[0]) * cos (k[1]) * cos (k[2]));
  }
  double result() {
    double res, err;

    double xl[3] = { 0, 0, 0 };
    double xu[3] = { M_PI, M_PI, M_PI };

    const gsl_rng_type *T;
    gsl_rng *r;

    ////// the following 3 lines didn't work ///////
    //function<double(A,double*, size_t, void*)> fg;
    //fg = &A::g;
    //gsl_monte_function G = { &fg, 3, 0 };
    gsl_monte_function G = { &g, 3, 0 };

    size_t calls = 500000;

    gsl_rng_env_setup ();

    T = gsl_rng_default;
    r = gsl_rng_alloc (T);

    {
      gsl_monte_plain_state *s = gsl_monte_plain_alloc (3);
      gsl_monte_plain_integrate (&G, xl, xu, 3, calls, r, s, &res, &err);
      gsl_monte_plain_free (s);
    }

    gsl_rng_free (r);
    return res;
  }
};

main() {
  A a;
  cout <<"gsl mc result is " << a.result() <<"
";
}

更新 (1):

我尝试将 gsl_monte_function G = { &g, 3, 0 }; 更改为 gsl_monte_function G = { bind(&A::g, this,_1,_2,_3), 3, 0 }; 但没用

I tried changing gsl_monte_function G = { &g, 3, 0 }; to gsl_monte_function G = { bind(&A::g, this,_1,_2,_3), 3, 0 }; but it didn't work

更新 (2):我尝试使用 将 std::function 分配给成员函数 但它没有也不能工作.

Update (2): I tried using assigning std::function to a member function but it didn't work either.

更新 (3)最后我写了一个非成员函数:

Update (3) in the end I wrote a non-member function:

double gmf (double *k, size_t dim, void *params) {
  auto *mf = static_cast<A*>(params);
  return abs(mf->g(k,dim,params));
  //return 1.0;
};

它有效，但这是一个混乱的解决方案，因为我需要编写一个辅助函数.使用 lambda、函数和绑定，应该有一种方法可以让类中的所有内容都合乎逻辑.

It worked but it's a messy solution because I needed to write a helper function. With lambdas,function and bind, there should be a way to have everything logical within the class.

推荐答案

您可以使用以下代码轻松包装成员函数(这是众所周知的解决方案)

You can easily wrap member functions using the following code (which is a well known solution)

 class gsl_function_pp : public gsl_function
 {
    public:
    gsl_function_pp(std::function<double(double)> const& func) : _func(func){
      function=&gsl_function_pp::invoke;
      params=this;
    }    
    private:
    std::function<double(double)> _func;
    static double invoke(double x, void *params) {
     return static_cast<gsl_function_pp*>(params)->_func(x);
   }
};

然后你可以使用 std::bind 将成员函数包装在一个 std::function 中.示例:

Then you can use std::bind to wrap the member function in a std::function. Example:

gsl_function_pp Fp( std::bind(&Class::member_function, &(*this),  std::placeholders::_1) );
gsl_function *F = static_cast<gsl_function*>(&Fp);

但是，在将成员函数包装在 gsl 集成例程中之前，您应该了解 std::function 的性能损失.参见 模板与 std::function .为避免这种性能下降(这对您来说可能重要也可能不重要)，您应该使用如下所示的模板

However, you should be aware about the performance penalties of std::function before wrapping member functions inside gsl integration routine. See template vs std::function . To avoid this performance hit (which may or may not be critical for you), you should use templates as shown below 

template< typename F >
  class gsl_function_pp : public gsl_function {
  public:
  gsl_function_pp(const F& func) : _func(func) {
    function = &gsl_function_pp::invoke;
    params=this;
  }
  private:
  const F& _func;
  static double invoke(double x, void *params) {
    return static_cast<gsl_function_pp*>(params)->_func(x);
  }
};

在这种情况下，要调用成员函数，您需要以下内容

In this case, to call a member function you need the following

 Class* ptr2 = this;
 auto ptr = [=](double x)->double{return ptr2->foo(x);};
 gsl_function_pp<decltype(ptr)> Fp(ptr);     
 gsl_function *F = static_cast<gsl_function*>(&Fp);

PS:链接 模板与 std::function 解释了编译器通常具有比 std::function 更容易优化模板(如果您的代码正在进行大量数值计算，这对性能至关重要).所以即使在第二个例子中的解决方法看起来更麻烦，我更喜欢模板而不是 std::function.

PS: the link template vs std::function explains that compiler usually has an easier time optimizing templates than std::function (which is critical for performance if your code is doing heavy numerical calculation). So even tough the workaround in the second example seems more cumbersome, I would prefer templates than std::function.

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