当模板类的输入输出类型不同时如何处理构造函数

Question

I'm learning C++ for only a short period of time and the following problem gives me headache.

What I'm trying to do is basiclly to wrap an existing library without introducing too much overhead such that the wrapper library could run as fast as the existing one. Therefore, I tend to not modify anything in existing library. I'm doing this in order to make the interface(syntax) compatible with my old codes.

Say, the existing class is called BASE, which is also templated class. There are a few ways to do the wrapping such as inheritance. I decided to go with the option of including BASE as a member of Wrapper class for better encapsulation.

        template<class T>    
        class Wrapper{
        public:
                     Wrapper() : base(){};
     /*error line*/  Wrapper( const Wrapper<double>& a, const Wrapper<double>& b ) : base(a.base, b.base){};
                    // constuct Wrapper<complex<double>> vector from two Wrapper<double> vectors using constuctor from existing class BASE
                    // 'a', 'b' are real and imag part respectively.

             private:
                     BASE<T>   base;
            }; 

The following line couldn't compile, error message is 'base is declared as private within the context'

      /*error line*/  Wrapper( const Wrapper<double>& a, const Wrapper<double>& b ) : base(a.base, b.base){};

Experiments I have done so far:

<1>. change

      private:
      BASE<T>   base;

to

      public:
      BASE<T>   base;

the code complies and gives correct answer. But, as I stated above, I want data encapsulation, so this solution is a no go.

<2>. Although the error message suggests there is something to do with the access privilege, I think this is caused by different input-output type (inputs are two "double", output is type of "complex double"). The following dosomething() function works as long as the type of input and type of (*this) are consistent, no error regarding 'base is declared as private within the context'.

    template<class T>    
    class Wrapper{
    public:
                 Wrapper() : base(){};
 /*error line*/  Wrapper( const Wrapper<double>& a, const Wrapper<double>& b ) : base(a.base, b.base){};

     void dosomething(const Wrapper<T>& a)
     {
      (*this).base = a.base; // ok, compiles good
     }

         private:
                 BASE<T>   base;
        }; 

How to work around? Look forwarding to any useful comments.

Answer 1

You may make your class friend of the same class but with different parameter:

template<class T>    
class Wrapper{
    template <typename U> friend class Wrapper;
public:
    Wrapper() : base(){}
    Wrapper(const Wrapper<double>& a, const Wrapper<double>& b) : base(a.base, b.base){}

private:
    BASE<T> base;
};