c++ - 将指向乘法类的指针传递给成员函数

Question

我有下一节课：

"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;
}

试图编译它我得到一个错误：

文件：integration.h，第 18 行，语法错误：'<tag>::*'

如何调用可能属于任何类的回调？

第二个问题：当我尝试在没有明确模板规范的情况下调用它时

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

我收到一个错误

文件：main.cpp，第 65 行，“MyIntegrator::integrate”：找不到匹配的重载函数

那么，为什么这个函数不能从它的参数中推断出它的参数呢？

在没有依赖默认参数的最后一个参数的情况下调用它时，我也会遇到同样的错误。

integrator_1.integrate(mp_id,span,step);

score 0 · Accepted Answer

用一点缩进解密你在这里的东西

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
}

看起来您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的声明，而不是一个方法。您需要使用指向方法的指针T。

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
}

但我不认为这将是犹太洁食，因为没有办法确保T并且Integrator以任何方式相关。

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解决的另一个问题，但它现在将编译并让我们继续。

但这不会

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

因为Integrator::base_integrate_callbackvoid Integrator::base_integrate_callback(const coord_type, double) 的签名与的签名不匹配void(caller::*callback)(const coord_type, double)。它们看起来一样，不是吗？缺少的是this所有方法都有的隐藏参数。caller::*callback期望 a caller *，但Integrator::base_integrate_callback提供Integrator *.

您可以通过make callerand it's ilk inheritIntegrator而不是来解决此问题MyIntegrator，但移至base_integrate_callback新的struct Integrated并拥有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，一个T为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);
}

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
}

全部一起：

#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::functionlambda表达式可以为您做什么。

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;
}

c++ - 将指向乘法类的指针传递给成员函数

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