在封闭类模板参数下初始化嵌套模板化类成员

Question

How to get around with this?

enum SID {SID_A, SID_B};

template<SID sid>
class Emp
{
    public:
    template<typename T> struct s{T a;};

    Emp<sid>(){
        if constexpr(sid == SID_A){
            s<T>(){a = 0;} //error: use of undeclared identifier 'T'
        }
        if constexpr(sid == SID_B){
            s<T>(){a = 1;} //error: use of undeclared identifier 'T'
        }
    }
};

Compiler: clang 9.0

Edit -----

I would like to achieve a default initialization of struct member sa depending on the argument of sid .
T would usually be an int or float .

Eg:

Emp<SID_A> emp_a; //here s.a = 0
Emp<SID_B> emp_b; //here s.a = 1

Answer 1

If I understand you correctly, you wish Emp have a data member, s , which is a struct containing a numerical data member, a . This member should be an int with initial value 0 if this is an Emp<SID_A> , whereas for an Emp<SID_B> , it should be a float with value 1.0. I am guessing you want to have the freedom to associate additional values of type SID with different numeric types and values for this->sa .

You can do this with template specialization. Declare the type of s as a struct S before the class Emp :

#include <type_traits>

enum SID {SID_A, SID_B};

template <SID> struct S;
template <> struct S<SID_A> { typedef int   Number; Number a;};
template <> struct S<SID_B> { typedef float Number; Number a;};
// You can define further specializations here

template<SID sid>
class Emp
{
  public:
    S<sid> s;

    Emp()
    {
      switch(sid)
      {
        case SID_A : s.a = 0.1; break; // Note value is 0.1 to prove it's an int
        case SID_B : s.a = 1.1; break; // Note value is 1.1 to prove it's a float
        // define further cases here
        default: ;
      }
    }
};

Now:

#include <iostream>

int main()
{
  Emp<SID_A> A;
  Emp<SID_B> B;
  std::cout << A.s.a << std::endl;
  std::cout << B.s.a << std::endl;
  return 0;
}

output:

0
1.1