将指针传递给乘法类到成员函数

时间:2016-09-11 14:13:59

标签: c++ templates member-pointers

我有下一个课程:

"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;
}

尝试编译它我收到一个错误:

  

file:integration.h,第18行,语法错误:'&lt;标签&GT; :: *'

如何调用可能属于任何类的回调?

第二个问题:当我尝试在没有像

这样的明确模板规范的情况下调用它时
integrator_1.integrate(mp_id,span,step,&A.callback);

我收到错误

  

file: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也会出现同样的错误。

integrator_1.integrate(mp_id,span,step);

1 个答案:

答案 0 :(得分:0)

通过一点缩进来解密你所拥有的东西

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指定默认值的方法。不幸的是,默认值看起来像另一个方法指针的声明,而不是方法。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
}

但我不认为这将是犹太人,因为无法确保TIntegrator以任何方式相关。

例如,清理后

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

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

因为您需要提供指向方法的指针,而不是指向方法的对象。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_callback,void Integrator :: base_integrate_callback(const coord_type,double)的签名与void(caller::*callback)(const coord_type, double)的签名不匹配。它们看起来一样,不是吗?缺少的是所有方法都隐藏的this参数。 caller::*callback期望caller *Integrator::base_integrate_callback提供Integrator *

您可以通过制作caller来解决这个问题,并继承Integrator而非MyIntegrator,但将base_integrate_callback移至新的struct Integratedcaller和朋友继承Integrated会更有意义。

回到我前面提到的另一个问题。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将需要一个参数,即T的引用,以便为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,因为上面将始终调用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
}

所有在一起:

#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:进入2011年,看看std::functionlambda expressions可以为您做些什么。

以下是一个例子:

#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;
}