我有一个派生自多个基类的类,我想使其成为一个单例。问题是派生类和基类不使用默认构造函数并接受参数,因此我对如何进行管理感到困惑。我希望能够将参数传递给构造函数,但是我只想做一次(我不希望它成为setter)。我唯一的解决方案是派生类的getInstance()成员中的静态布尔值。

基本情况:

//Singleton Derived Class with multiple bases of non-default constructors
class base1 {
    public:
        base1(int* value) :
              val_{value} {;}
        ~base1();
    private:
        int val_;
}

class base2 {
    public:
        base2(int* value) :
              val_{value} {;}
        ~base2();
    private:
        int val_;
}

class derived : public base1, public base2 {
    public:
        derived(int* value) :
                base1{value},   //Base 1 constructor call
                base2{value},   //Base 2 constructor call
                val_{value} {;}
        ~derived();
    private:
        int val_;
}

//Creation
derived newDerived(&value);


尝试使其变得单身吗?

//Lets make it a singleton
class base1 {
    public:
        base1(); //Can I pass the construtor anything?
        ~base1();
    private:
        int val_;
}

class base2 {
    public:
        base2(); //Can I pass the construtor anything?
        ~base2();
    private:
        int val_;
}

class derived : public base1, public base2 {
    public:
        static derived& getInstance(int* value) {
            static bool init;
            if (!init) {
                base1::val_ = value;
                base2::val_ = value;
                init=true;
            }
            static derived instance;
            return instance;
        }
        derived(int* value) {;}
        ~derived();
    private:
        derived(derived const&) = delete;           //Copy construct
        derived(derived&&) = delete;                //Move construct
        derived& operator=(derived const&) = delete;//Copy assign
        derived& operator=(derived &&) = delete;    //Move assign
        int val_;
}

//Creation
derived::getInstance(&value);


我正在寻找有关如何执行此操作的方向,或者也许是我根本不应该执行此操作的原因?谢谢

最佳答案

干得好。而不是参考,我在这里使用指针。这只是一个示例,因此我不能使此单例成为线程安全的。还要将默认构造函数声明为私有,这样就无法实例化。

class base1 {
public:
     base1(int* value) : val_(*value) {}
    ~base1() {};
    private:
        int val_;
};

class base2 {
 public:
    base2(int* value) :
    val_(*value ) { }
    ~base2() {};
private:
    int val_;
};

class derived : public base1, public base2 {
    private:
    derived(int* value) :
    base1(value),   //Base 1 constructor call
    base2(value),   //Base 2 constructor call
    val_(*value) {}
   ~derived() {};
     private:
     int val_;
     static derived* driv;

    public:
  static derived* getInstance(int* value);
};

 derived* derived::driv = NULL;

derived* derived::getInstance(int* value)
{
    if (driv == NULL)
    {
      driv = new derived(value);
    }
  return driv;
}

关于c++ - 具有非默认构造函数的多个基的Singleton派生类,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/47023341/

10-11 19:35
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