c++ - C++ 中的模板和继承

Question

我正在尝试按如下方式设置类层次结构:

template< class TProcess>
class Path
{
protected:
    TProcess &process;
    double Xt;
public:
    Path(TProcess &process, double X0) : process(process), Xt(X0) {}
    virtual ~Path(){}

    virtual double dX(double dt, const std::vector<double> &dW) = 0;
};

template< class TProcess>
class Process
{
protected:
    double X0;
public:
    Process(double X0) : X0(X0) {}
    virtual ~Process(){}

    virtual boost::shared_ptr<Path<TProcess> > NewPath() =0;
};

但后来我发现无法声明子类以使其工作。我试过:

class GeometricBrownianMotion;
class GBMPath: public Path<GeometricBrownianMotion>
{
    double dX(double dt, const std::vector<double> &dW)
    {
        Xt = Xt*exp(process.mu*dt - 0.5*dt + process.sigma*dW);
        return Xt;
    }
};

class GeometricBrownianMotion: public Process<GeometricBrownianMotion>
{
    double mu;
    const std::vector<double> &sigma;
    double sigma2;
public:
    GeometricBrownianMotion(double X0, double mu, const std::vector<double> &sigma): Process(X0), mu(mu), sigma(sigma), sigma2(sigma*sigma)
    {}

    virtual boost::shared_ptr<Path<GeometricBrownianMotion> > NewPath()
    {
        return boost::shared_ptr<Path<GeometricBrownianMotion> >(new GBMPath(*this, X0));
    }

};

但我收到以下错误:

process.h:58: error: forward declaration of ‘struct GeometricBrownianMotion’
process.h:63: error: invalid use of incomplete type ‘struct GeometricBrownianMotion’

有什么想法可以让它发挥作用吗？

Answer 1

只需将您的实现与您的类定义分开，问题就会自动消失。

几乎是这样，因为您可能只想移动 GeometricBrownianMotion 的定义，然后再将其用作模板参数。

编辑:即使差一点，至少在我的编译器上，删除所有内联方法定义就足以让它编译。<​​/p>