C++簡單的單例實現返回語句的構造函數

Question

我想在C++中實現一個簡單的單例模式。

#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; 
} 

我的問題是關於多個構造函數可供simpleSingleton類

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

構造函數應返回的類的對象。在上面的例子中，這兩個構造函數都沒有返回語句。然而這似乎是標準的簡單實現。這是爲什麼？

Answer 1

如果你用單，用邁耶斯的單例：

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; 
    } 
};