c passing template param by shared pointer-CodePudding

wanna ask one question: I have two classes A and B, few data types can only be defined in class B, but A as a higher level class also need use these data types defined in class B. So, I defined a template function in class A, named define_spline(), I can using robot_1 (i.e., class B) class's pointer to passing the type, but what I want is the class A can have a member object of these types defined in B class (robot_1 struct sp{}). I try to write a template class for whole A, named A2, but I got error 'rp1 is not a type name', why this is ok in class A define_spline() function, but not work for class A2 ? and what's the correct way to do so?

#include<Eigen/Dense>
#include<iostream>
#include <unsupported/Eigen/Splines>
#include <fstream>
#include<memory>

class robot_1
{
    private: 
        static const int nState =9;
        static const int nControl =3;

    public:

        // robot_1(int nx, int nu)
        //     printf("robot_1 constructor \n");
        // }
        ~robot_1(){}

        robot_1()
        {
             printf("empty robot_1 constructor \n");
        }


        struct sp
        {   
            using SplineX = Eigen::Spline<double, nState>;
            using SplineFittingX = Eigen::SplineFitting<SplineX>;

        }sp_;
};

using robotPtr = std::shared_ptr<robot_1>;



class A
{
    private:
        std::string name = "class_a";
        int nState;
        int ncontrol;
        
        // Ty::SplineX sp_memeber;


    public:
        A(){}
        ~A(){}
 
       template <typename Ty>
        void  define_spline(const Ty, int a)
        {
            using SplineX_A = typename Ty::SplineX;
            using SplineFittingX_A =  typename Ty::SplineFittingX;

            SplineX_A  spA; 
            std::cout<<a<<std::endl;
        }

        robotPtr robotPtr_;

};



template <typename Type2>
class A2
{
    private:
       Type2::SplineX spx;
       Type2::SplineFittingX spxF;
        int nState;
        int nControl;
    public:
        A2(){}
        ~A2(){}
        A2(int nx, int nu){
            nState = nx;
            nControl = nu;
        }

};



int main()
{
    robotPtr rp1 = std::make_shared<robot_1>();
    rp1 ->sp_;

    A a;
    a.robotPtr_ = std::make_shared<robot_1>();

    a.define_spline(rp1->sp_, 6); // here no error

    // A2
    A2<rp1->sp_> a2(9,3); // error

}

Thanks in advance

CodePudding user response：

A2<rp1->sp_> a2(9,3); // error

Template parameters are always types (until very recently, but this is not material right now). rt1->sp_ is not a type. It is a discrete object. If you look where it appears, struct sp is the type in question. And sp_ is an instance of that type. Types, and objects, are two completely different things, in C . They are not interchangeable with each other. When something must be a type, it has to be a type, and not an object. When something must be an object it has to be an object, and not a type. Template parameters are types (until very recently, but this is still not material right now).

You might get this to work by using A2<decltype(rp1->sp_)>. The expression decltype(...) resolves to whatever the reference type is. It basically "translates" an object, of some kind, to its type. There are a few rules, and several nuanced rules to decltype(), that your favorite C textbook will be happy to explain to you, which is where you can look for more information and examples.