没有默认构造函数的模板的构造函数

Question

I have a template class that currently has a constructor that takes no arguments. The issue is that in some cases the class that is being used in the template does not have an empty constructor which gives compile errors.

template <typename T>
class A
{
     public:
     T Thing;
     int number;
     A():number(5) {}
 };


  class B
  {
       public:
       int a;
       B(int _a):a(_a) {}
  }

  A<int> a1; // This is fine
  A<B> a2; // This is not fine since B has no default constructor

I've thought of possibly using adding a second constructor which takes a ref to T, and then using enable_if to remove the parameterless constructor if T does not have a default constructor, but I'm not sure if that would work. Are there other options?

Note that I do not want to add a default constructor to class B because I don't want to leave an instance of B in an unknown state.

Answer 1

Spoke to some people and did more research and found the solution. Making the default constructor templated by a constexpr static member. Here it is:

template <typename T>
class A
{
public:
     T Thing;
     int number;
     static constexpr bool HasDefaultConstructors = std::is_default_constructible<T>;

     // This removes the default constructor if T does not have one
     template<bool HDC = HasDefaultConstructors, typename std::enable_if<HDC, int>::type = 0>
     A():number(5) {}

     A(const Thing &t):T(t) {}
 };