Question

我正在尝试用C ++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simpleSingleton类

可用的多个构造函数

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

构造函数应该返回该类的对象。在上面的示例中，两个构造函数都没有return语句。然而，这似乎是标准的简单实现。那是为什么？

Answer 1

如果你使用单身人士，请使用迈耶斯的单身人士：

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;
    }
};

构造函数的C ++简单单例实现返回语句

1 个答案: