one class should invoke method of another class us

2019-08-03 08:41发布

I have consult (disclaimer):

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;
}

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);

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.

1条回答
爷、活的狠高调
2楼-- · 2019-08-03 09:02

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;

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

Edit:

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);

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

查看更多
登录 后发表回答