一个类应该使用函数指针来调用另一类的方法 [英] one class should invoke method of another class using a function pointer

问题描述

我已咨询(免责声明):

I have consult (disclaimer):

• 类成员函数指针
• 通过函数指针调用C ++类方法
• C ++:指向另一个类函数的函数指针

• Class member function pointer
• Calling C++ class methods via a function pointer
• C++: Function pointer to another class function

为说明我的问题，我将使用以下代码(已更新)

To illustrate my problem I will use this code (UPDATED)

#include <iostream>
#include <cmath>
#include <functional>


class Solver{
public:
    int Optimize(const std::function<double(double,double>& function_to_optimize),double start_val, double &opt_val){
        opt_val = function_to_optimize(start_val);
        return 0;
    }
};

class FunctionExample{
public:
    double Value(double x,double y)
    {
        return x+y;
    }
};

int main(){

    FunctionExample F =FunctionExample();
    Solver MySolver=Solver();

    double global_opt=0.0;

    MySolver.Optimize(std::bind(&FunctionExample::Value, &F, std::placeholders::_2),1,global_opt);

    return 0;
}

有没有一种方法可以调用方法"Value"?我没有问题可以调用一个函数(没有类)

Is there a way to call the method "Value"? I have no problems to call a function (without a class)

typedef double (*FunctionValuePtr)(double x);

但这对上面的示例没有帮助.我需要显式方法名称.大多数示例使用静态方法.我不能使用静态方法.

But this does not help me with the example above. I need the explicit method name. Most examples use a static method. I can not use a static method.

推荐答案

您可以使用STL的<functional>标头:

You can use the <functional> header of the STL:

double Gradient(const std::function<double(double)>& func, double y)
{
    const double h = 1e-5;
    return (func(y+h) - func(y)) / h;
}

std::cout << D.Gradient(std::bind(&Root::Value, &R, std::placeholders::_1), 8) << std::endl;

就像乔亚希姆·皮尔博格(Joachim Pileborg)所说的那样，您在main中声明了函数，因此您需要删除().

Also like Joachim Pileborg commented you are declaring functions in main, so you need to remove the ().

要为bind提供一个固定的参数，您可以执行以下操作:

To give bind a fixed argument you can do the following:

int Optimize(const std::function<double(double)>& function_to_optimize, double &opt_val){
    opt_val = function_to_optimize(opt_val);
    return 0;
}

MySolver.Optimize(std::bind(&FunctionExample::Value, &F, std::placeholders::_1, 1), global_opt);

这将称为F.Value(opt_val, 1).您还可以将占位符与固定参数互换.

This will call F.Value(opt_val, 1). You can also swap the placeholder with the fixed argument.

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