重載運營商的std :: ostream的和運營商<<打印實例內存地址

Question

我想超載std::ostream& operator<<這樣做：

#include <iostream> 

class MyTime 
{ 
private: 
    unsigned char _hour; 
    unsigned char _minute; 
    float _second; 
    signed char _timeZone; 
    unsigned char _daylightSavings; 
    double _decimalHour; 

    friend std::ostream& operator<<(std::ostream&, const MyTime&); 
public: 
    MyTime(); 
    ~MyTime(); 
}; 

和操作執行：

std::ostream& operator<<(std::ostream &strm, const MyTime &time) 
{ 
    return strm << (int)time._hour << ":" << (int)time._minute << ":" << time._second << " +" << (int)time._daylightSavings << " zone: " << (int)time._timeZone << " decimal hour: " << time._decimalHour; 
} 

但是，當我這樣做：

MyTime* anotherTime = new MyTime(6, 31, 27, 0, 0); 
std::cout << "Time: " << anotherTime << std::endl; 

它只打印anotherTime內存地址。

我在做什麼錯？

Answer 1

您的意思是：

std::cout << "Time: " << *anotherTime << std::endl; 

，或者甚至更好：

MyTime anotherTime (6, 31, 27, 0, 0); 
std::cout << "Time: " << anotherTime << std::endl; 

Answer 2

你的運營商期望通過const引用一個實例對象，但你給它一個指針。要麼是std::cout << "Time: " << *anotherTime << std::endl;（注意*）或添加其他運營商：

friend std::ostream& operator<<(std::ostream&, const MyTime*);

Answer 3

std::cout << "Time: " << anotherTime << std::endl; 

以上anotherTime實例代碼

是指明MyTime對應*，但必須指明MyTime對應。

你有兩個解決方案來解決這個問題：

製作堆棧指明MyTime對應，並且它會正常工作，因爲現在anotherTime不是指針。

MyTime anotherTime(6, 31, 27, 0, 0); 
    std::cout << "Time: " << anotherTime << std::endl; 

間接引用代碼中的指針，就像這樣：

MyTime* anotherTime = new MyTime(6, 31, 27, 0, 0); 
std::cout << "Time: " << *anotherTime << std::endl;