在单例类中使用智能指针时出现C2248错误

Question

在下面的代码中我遇到了 C2248 错误。当我用原始指针替换智能指针时，它会编译。不过我想使用智能指针 std::shared_ptr<>目的。 使用智能指针时如何纠正此错误？

    #include <iostream>
    #include <string>
    #include <memory>

    class GameSetting{
    inline static std::shared_ptr<GameSetting> _instance;
    unsigned int _brightness;
    unsigned int _width;
    unsigned int _heigth;
    GameSetting():_brightness{150}, _width{300}, _heigth{200}
   {}

   public:
   static std::shared_ptr<GameSetting> getInstance(){
    if(! _instance)
        _instance=std::make_shared<GameSetting>();
    return _instance;
   }


  void setBrightness(unsigned brightness){_brightness=brightness;}
  void setWidth(unsigned  int width){_width=width;}
  void setHeight(unsigned int height){_heigth=height;}

  unsigned int getBrightness()const{return  _brightness;}
  unsigned int getWidth()const{return _width;}
  unsigned int getHeigth()const{return _heigth;}

  void displaySettings()const{
    std::cout<<"brigthness: "<<_brightness<<'\n'<<"width: "<<_width
            <<'\n'<<"heigth: "<<_heigth<<"\n\n";
  }
  };



  int main()
  {
  std::shared_ptr<GameSetting> setting=GameSetting::getInstance();
  setting->displaySettings();

  }

Answer 1

make_shared() 调用需要访问 GameSetting 类的默认构造函数；但是，该构造函数被隐式声明为private（这是类成员的默认访问说明符）。

要解决此问题，只需在构造函数定义之前添加 public: 访问说明符即可：

class GameSetting {
    inline static std::shared_ptr<GameSetting> _instance;
    unsigned int _brightness;
    unsigned int _width;
    unsigned int _heigth;
public: // Anything before this is "private" by default
    GameSetting() :_brightness{ 150 }, _width{ 300 }, _heigth{ 200 }
    {}
public:
    static std::shared_ptr<GameSetting> getInstance()
    {
        if (!_instance)
            _instance = std::make_shared<GameSetting>();
        return _instance;
    }
   // ... the rest of you class definition

或者，如果您希望/需要将 GameSetting 构造函数保持私有（与单例中常见的情况相同）类），那么您可以将对 std::make_shared 的调用替换为对 std::shared_ptr 构造函数的直接“调用”（以及结果的赋值）。因此，您的 getInstance 函数将是：

    static std::shared_ptr<GameSetting> getInstance()
    {
        if (!_instance)
            _instance = std::shared_ptr<GameSetting>{new GameSetting};
        return _instance;
    }