C++简单的单例实现返回语句的构造函数

Question

我想在C++中实现一个简单的单例模式。

#include <iostream> 

class simpleSingleton 
{ 
private: 
    static simpleSingleton * pInstance; 
    simpleSingleton(){} 
    simpleSingleton(const simpleSingleton& rs) { 
     pInstance = rs.pInstance; 
    } 
    simpleSingleton& operator = (const simpleSingleton& rs) 
    { 
     if (this != &rs) 
     { 
      pInstance = rs.pInstance; 
     } 
     return *this; 
    } 
    ~simpleSingleton(){}; 

public: 
    static simpleSingleton& getInstance() 
    { 
     static simpleSingleton theInstance; 
     pInstance = &theInstance; 

     return *pInstance; 
    } 
    void demo() 
    { 
     std::cout << "simple singleton." 
        << std::endl; 
    } 
}; 
simpleSingleton *simpleSingleton::pInstance = nullptr; 


int main() 
{ 
    /*Version 1 */ 
    simpleSingleton * p = &simpleSingleton::getInstance(); // cache instance pointer p->demo(); 
    p->demo(); 

    /*Version 2 */ 
    simpleSingleton::getInstance().demo(); 

    return 0; 
} 

我的问题是关于多个构造函数可供simpleSingleton类

simpleSingleton(){}' 'simpleSingleton(const simpleSingleton& rs)

构造函数应返回的类的对象。在上面的例子中，这两个构造函数都没有返回语句。然而这似乎是标准的简单实现。这是为什么？

Answer 1

如果你用单，用迈耶斯的单例：

class Singleton 
{ 
private: 
    Singleton() = default; // Adapt here 
    ~Singleton() = default; // Adapt here 
    Singleton(const simpleSingleton& rs) = delete; 
    Singleton& operator = (const simpleSingleton&) = delete; 
public: 
    Singleton& GetInstance() 
    { 
     static Singleton instance; 
     return instance; 
    } 
};