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如何使用動態綁定覆蓋純虛擬 function

[英]How to overwrite a pure virtual function with dynamic binding

我正在嘗試解決 Pitt-Francis 文本https://link.springer.com/book/10.1007/978-3-319-73132-2中關於抽象 ODE class 的練習 7.3。

抽象 class 給出為

class AbstractODESolver
{
private:
    ...
    
public:
    ....
    virtual double RightHandSide(double y, double t) = 0;
    virtual double SolveEquation() = 0;
    
};

派生的 class 給出為

class ForwardEuler: public AbstractODESolver
{
public:
     double RightHandSide(double y, double t);
     double SolveEquation();
};

如何在 ForwardEuler 中轉換 RightHandSide 以便在運行時定義它?

特別是主文件可以采用以下形式

double RHS(double y, double t);

int main(int argc, char* argv[])
{
    
    AbstractODESolver* p_ODEs = new ForwardEuler;
        
    return 0;
}

double RHS(double y, double t)
{
    return 1+t;
}

我想寫一些類似的東西

p_ODEs->RightHandSide = &RHS;` 

指向 function RHS

ForwardEuler class 一個數據成員來存儲 function,並在構造函數中傳遞它。 RightHandSide function 只會調用傳入的 function。

class ForwardEuler: public AbstractODESolver
{
public:
    ForwardEuler(std::function<double(double, double)> rhs): rhs(rhs) {}
    double RightHandSide(double y, double t) { return rhs(y, t); }
    double SolveEquation();

private:
    std::function<double(double, double)> rhs;
};

...

double RHS(double y, double t);
AbstractODESolver* p_ODEs = new ForwardEuler(RHS);

這是我不需要std::function的解決方案,我認為這是本章教授的技術所需的解決方案。

(根據@Paul Sanders)

class ForwardEuler: public AbstractODESolver
{
public:
     ForwardEuler(double (*pRhs)(double, double));
     double RightHandSide(double y, double t);
     double SolveEquation();

private:
     double (*mRhs)(double y, double t);     
};
ForwardEuler::ForwardEuler(double (*pRhs)(double, double))
{
    mRhs = pRhs;
}

double ForwardEuler::RightHandSide(double y,double t)
{
    return (*mRhs)(y,t);
}

主function

double RHS(double y, double t);

int main(int argc, char* argv[])
{
    
    AbstractODESolver* p_ODEs = new ForwardEuler(RHS);
    std::cout<<(*p_ODEs).RightHandSide(1.0,1.0)<<std::endl;

    return 0;
}

double RHS(double y, double t)
{
    return 1+t;
}

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