template<class Stepper>替换所需的

Question

当我尝试编译以下代码时，会遇到错误

在“struct result_of_make_controlled >”的实例化中： 54:53:将“template typename result_of_make_controlled::type make_controlled(typename Stepper::value_type，typename Stepper::value_type，const Stepper&)”替换为Stepper =runge_kutta_dopri5所需 69:60:需要从这里开始 49:54:错误:在‘struct get_controller >’中没有名为“type”的类型 类型名称get_controller<步骤>：：类型类型；

class explicit_error_stepper_fsal_base 
{
public:
typedef double state_type;
typedef double value_type;

};


template<class State>
class runge_kutta_dopri5
: public explicit_error_stepper_fsal_base
{

public :

    typedef explicit_error_stepper_fsal_base stepper_base_type;

    typedef typename stepper_base_type::value_type value_type;//#

    runge_kutta_dopri5(  ) 
    { }

};


template< class Stepper > struct get_controller { };



// default controller factory
template< class Stepper , class Controller >
struct controller_factory
{
    Controller operator()(
            typename Stepper::value_type abs_error ,
            typename Stepper::value_type rel_error ,
            const Stepper &stepper )
    {
        return Controller( abs_error , rel_error , stepper );
    }
};


template< class Stepper >
struct result_of_make_controlled
{
    typedef typename get_controller< Stepper >::type type;
};


template< class Stepper >
typename result_of_make_controlled< Stepper >::type make_controlled(
        typename Stepper::value_type abs_error ,
        typename Stepper::value_type rel_error ,
        const Stepper & stepper = Stepper() )
{
    typedef Stepper stepper_type;
    typedef typename result_of_make_controlled< stepper_type >::type controller_type;
    typedef controller_factory< stepper_type , controller_type > factory_type;
    factory_type factory;
    return factory( abs_error , rel_error , stepper );
}


typedef double state_type;
typedef runge_kutta_dopri5<state_type> stepper_type;
typedef decltype(make_controlled(1E-10,1E-10,stepper_type())) controlled_stepper_type;


int main()
{
    return 0;
}

我假设它需要在结构type中查找一个名为get_controller的类型，而它是空的，并且会导致错误。我不明白的是，为什么在github boost库的原始源代码中，编译器方面没有问题？

Answer 1

相关的boost报头为特定的步进者提供了get_controller的部分专门化。您的代码不会这样做。例如

template< class State , class Value , class Deriv , class Time , class Algebra , class Operations , class Resize >
struct get_controller< runge_kutta_dopri5< State , Value , Deriv , Time , Algebra , Operations , Resize > >
{
    typedef runge_kutta_dopri5< State , Value , Deriv , Time , Algebra , Operations , Resize > stepper_type;
    typedef controlled_runge_kutta< stepper_type > type;
};