传递指向 multiply 类的指针给成员函数

Question

我有下一个 classes:

"Integrator.h"

#include <vector>
#include <array>
using namespace std;

class Integrator {
public:
    using coord_type = array<double, 3>;  
protected:
    void base_integrate_callback(const coord_type, double t_k) {
      //does nothing
    }
};

class MyIntegrator :public Integrator {
public:
   template <class T>
   void integrate(int mp_id, int t_span, int step ,
   void(T::*callback)(const coord_type, double) = (Integrator::*)(const coord_type, double)){
  //calls callback here
}
};

"main.cpp"

#include Integrator.h"

struct caller {
   void callback(const Integrator::coord_type coord, double t_k) {
   //does smth
}
};

int main(){
   MyIntegrator integrator_1;
   caller A;
   int mp_id = 1;
   int span = 365;
   int step = 1;
   integrator_1.integrate<caller>(mp_id,span,step,&A.callback);
   return 0;
}

尝试编译时出现错误：

file:integration.h, line 18, syntax error: '< tag>::*'

如何调用属于任何 class 的回调？

第二个问题：当我尝试在没有显式模板规范的情况下调用它时，例如

integrator_1.integrate(mp_id,span,step,&A.callback);

我收到一个错误

file: main.cpp , line 65, 'MyIntegrator::integrate': no matching overloaded function found

那么，为什么这个函数不能从它的参数推导出它的参数？

我在没有依赖默认参数的最后一个参数调用它时也得到同样的错误。

integrator_1.integrate(mp_id,span,step);

Answer 1

用一点缩进解密你在这里的内容

template <class T>
void integrate(int mp_id, 
               int t_span, 
               int step ,
               void(T::*callback)(const coord_type, double) = (Integrator::*)(const coord_type, double))
{
    //calls callback here
}

您似乎正在尝试声明一个将回调函数作为参数并分配默认值的方法。不幸的是，默认值看起来像是另一个方法指针而不是方法的声明。您需要使用指向 T.

方法的指针

template <class T>
void integrate(int mp_id, 
               int t_span, 
               int step ,
               void(T::*callback)(const coord_type, double) = &Integrator::base_integrate_callback)
{
    //calls callback here
}

但我认为这不会是犹太洁食，因为无法确保 T 和 Integrator 有任何关联。

例如清理后

integrator_1.integrate < caller > (mp_id, span, step, &A.callback);

到

integrator_1.integrate < caller > (mp_id, span, step, &caller::callback);

因为您需要提供指向方法的指针，而不是引用方法的对象。这暴露了我们稍后会遇到的另一个问题，但它现在会编译并让我们继续。

但这不会

integrator_1.integrate < caller > (mp_id, span, step);

因为Integrator::base_integrate_callback的签名，void Integrator::base_integrate_callback(const coord_type,double)，与void(caller::*callback)(const coord_type, double)的签名不匹配。他们看起来一样，不是吗？缺少的是隐藏的 this 参数，所有方法都有。 caller::*callback 期望 caller *，但 Integrator::base_integrate_callback 提供 Integrator *。

您可以通过使 caller 继承 Integrator 而不是 MyIntegrator 来解决此问题，但将 base_integrate_callback 移动到新的 struct Integrated 并具有caller 和朋友继承 Integrated 会更有意义。

回到我之前提到的另一个问题。在

template <class T>
void integrate(int mp_id, 
               int t_span, 
               int step ,
               void(T::*callback)(const coord_type, double) = &Integrated::base_integrate_callback)
{
    coord_type x; // junk for example
    double y; //junk for example
    callback(x,y); //KABOOM!
}

在什么对象上调用回调？ integrate 还需要一个参数，即对 T 的引用以为 callback 提供上下文。

template <class T>
void integrate(int mp_id, 
               int t_span, 
               int step,
               T & integrated,
               void(T::*callback)(const coord_type, double) = &Integrated::base_integrate_callback)
{
    coord_type x; // junk for example
    double y; //junk for example
    integrated.callback(x,y);
}

然后你必须使用正确的语法来调用函数指针，因为上面总是调用caller::callback。

template <class T>
void integrate(int mp_id, 
               int t_span, 
               int step,
               T & integrated,
               void(T::*callback)(const coord_type, double) = &Integrated::base_integrate_callback)
{
    coord_type x; // junk for example
    double y; //junk for example
    (integrated.*callback)(x,y); //std::invoke would be preferred if available
}

总计：

#include <array>
#include <iostream>

class Integrator
{
public:
    using coord_type = std::array<double, 3>;
};

struct Integrated
{
    void base_integrate_callback(const Integrator::coord_type, double t_k)
    {
        std::cout << "made it to default" << std::endl;
    }
};

class MyIntegrator: public Integrator
{
public:
    template <class T>
    void integrate(int mp_id,
                   int t_span,
                   int step,
                   T & integrated,
            void(T::*callback)(const coord_type, double) = &Integrated::base_integrate_callback)
    {
        coord_type x; // junk for example
        double y = 0; //junk for example
        (integrated.*callback)(x,y);
    }
};


struct caller:public Integrated
{
    char val; // for test purposes
    caller(char inval): val(inval) // for test purposes
    {

    }
    void callback(const Integrator::coord_type coord, double t_k)
    {
        std::cout << "made it to " << val << std::endl;
    }
};

int main()
{
    MyIntegrator integrator_1;
    caller A {'A'};
    caller B {'B'};
    caller C {'C'};
    int mp_id = 1;
    int span = 365;
    int step = 1;
    integrator_1.integrate < caller > (mp_id, span, step, A, &caller::callback);
    integrator_1.integrate < caller > (mp_id, span, step, B, &caller::callback);
    integrator_1.integrate < caller > (mp_id, span, step, C);
    return 0;
}

建议：步入 2011 年，看看 std::function and lambda expressions 能为您做些什么。

这是一个例子：

#include <array>
#include <iostream>
#include <functional>

class Integrator
{
public:
    using coord_type = std::array<double, 3>;
};

// no need for integrated to get default callback

class MyIntegrator: public Integrator
{
public:
    template <class T>
    void integrate(int mp_id,
                   int t_span,
                   int step,
                   // no need to provide object instance for callback. packed with std::bind
                   std::function<void(const coord_type, double)> callback =
                           [](const coord_type, double) { std::cout << "made it to default" << std::endl; })
                           // default callback is now lambda expression
    {
        coord_type x; // junk for example
        double y = 0; //junk for example
        callback(x,y); // no weird syntax. Just call a function
    }
};


struct caller
{
    char val; // for test purposes
    // no need for test constructor
    void callback(const Integrator::coord_type coord, double t_k)
    {
        std::cout << "made it to " << val << std::endl;
    }
};

int main()
{
    MyIntegrator integrator_1;
    caller A {'A'};
    caller B {'B'};
    // no need for test object C
    int mp_id = 1;
    int span = 365;
    int step = 1;
    using namespace std::placeholders; // shorten placeholder names
    integrator_1.integrate < caller > (mp_id, 
                                       span, 
                                       step, 
                                       std::bind(&caller::callback, A, _1, _2));
    // std bind bundles the object and the callback together into one callable package

    integrator_1.integrate < caller > (mp_id, 
                                       span, 
                                       step, 
                                       [B](const Integrator::coord_type p1, 
                                           double p2) mutable // lambda captures default to const 
                                       { 
                                           B.callback(p1, p2); // and callback is not a const method
                                       });
    // Using lambda in place of std::bind. Bit bulkier, but often swifter and no 
    //need for placeholders

    integrator_1.integrate < caller > (mp_id,
                                       span,
                                       step,
                                       [](const Integrator::coord_type p1,
                                           double p2)
                                       {
                                           std::cout << "Raw Lambda. No callback object at all." << std::endl;
                                       });
    //custom callback without a callback object

    integrator_1.integrate < caller > (mp_id, span, step);
    //call default

    return 0;
}

传递指向 multiply 类的指针给成员函数

passing pointer to multiply classes to a member function

c++

templates

member-pointers

传递指向 multiply 类 的指针给成员函数

passing pointer to multiply classes to a member function

c++

templates

member-pointers

传递指向 multiply 类的指针给成员函数