是否可以在 C++ 中使用非默认构造函数创建模板实例？

Question

从基于模板的类继承时是否可以指定使用哪个构造函数？这似乎是一个基本的事情，但我似乎无法弄清楚如何在 C++ 中实现这一点......

在下面的代码中，我想在创建 PlayerActor（继承自 BaseActor 模板类）时对 PlayerStats 对象使用非默认构造函数。

class BaseStats {
public:
   BaseStats ()
   {

   }

private:
};

class MonsterStats : public BaseStats {
public:
   MonsterStats()
   {

   }

private:
};

class PlayerStats : public BaseStats {
public:
   PlayerStats(
      const bool is_new) : 
         m_is_new(is_new)
   {

   }

private:
   const bool m_is_new;
   PlayerStats(); //Don't want this being used...
};

template <class ActorStatsClass>
class BaseActor
{
public:
   BaseActor()
   {

   }

private:
   ActorStatsClass m_stats;
};

class MonsterActor: public BaseActor<MonsterStats> {
public:
   MonsterActor()
   {

   }
private:
};

// This is where I'm not sure how to tell the template to use a non-default constructor...
// I get an error saying "PlayerStats::PlayerStats": cannot access private member declared in "PlayerStats". (Which is intended, but I don't want it to try and use the default constructor...)
class PlayerActor : public BaseActor<PlayerStats> {
public:
   PlayerActor(const bool is_new)
   {

   }
private:
   PlayerActor(); // Don't use this...
};

Answer 1

template <class ActorStatsClass>
class BaseActor
{
public:
   BaseActor()
   {

   }

private:
   ActorStatsClass m_stats;
};

如果此类BaseActor将具有任意成员，则它需要构造这些任意成员的方法。所以它需要一个通用的构造函数。

template <class ActorStatsClass>
class BaseActor
{
public:
   template<class...Us>
   BaseActor(Us&&...vs)
   : m_stats(std::forward<Us>(vs)...)
   { }

private:
   ActorStatsClass m_stats;
};

这使用转发，无论参数传递给BaseActor，它只是将它们直接传递到构造m_stats成员。

这允许：

class PlayerActor : public BaseActor<PlayerStats> {
public:
   PlayerActor(const bool is_new)
   : BaseActor<PlayerStats>(is_new)
   { }
   PlayerActor() = delete; // Note: this is better than making it private
};