Question

我收到了这个警告，它在整个构建过程中重复出现警告来自以下代码逻辑：行号6

class sample
{
private:                                                               //1
    // Private so that it can not be called                            //2
    sample();                                                          //3
    sample(sample const&);                                             //4
    // Assigning copy constructor with default constructor             //5
    sample& operator= (sample const&){};                               //6
    //                                                                 //7
    static sample *m_Instancesample;                                   //8
    static bool m_binstanceFlagsample;                                 //9
public:                                                                //10
    static sample *getInstance();                                      //11
    ~sample();                                                         //12
    void dummy(); 
};

Answer 1

您的sample& operator= (sample const&){};已定义了返回类型，但您没有返回任何内容。实现该功能或删除{}

Answer 2

sample& operator= (sample const&){};  
                                 ^^

{}部分用空体定义函数 - 定义缺少return语句。

定义函数正确或删除{}（即声明函数并在别处定义）。

Answer 3

替换如下：

  sample& operator= (sample const&){};

使用：

  const sample& operator= (sample const& sampleInstance)
    {
        //Get the instance of your sample class as it is singleton you will get the same instance 
        return *sample::getInstance();
    };

警告＃1011：在非void函数结束时缺少return语句

3 个答案: