[英]Overloaded operator std::ostream& operator<< prints instance memory address
我正在尝试重载std::ostream& operator<<
执行此操作:
#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();
};
和运营商的实施:
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;
}
但是当我这样做时:
MyTime* anotherTime = new MyTime(6, 31, 27, 0, 0);
std::cout << "Time: " << anotherTime << std::endl;
它只打印anotherTime
内存地址。
我究竟做错了什么?
你的意思是:
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;
您的操作符需要const引用的实例对象,但是您给它一个指针。 要么做std::cout << "Time: " << *anotherTime << std::endl;
(注意*
)或添加另一个运算符:
friend std::ostream& operator<<(std::ostream&, const MyTime*);
std::cout << "Time: " << anotherTime << std::endl;
在上面的代码中,anotherTime实例是myTime *,但它必须是myTime。
你有两个解决方案来解决这个问题:
在堆栈上制作myTime,它会正常工作,因为anotherTime现在不是指针。
MyTime anotherTime(6, 31, 27, 0, 0);
std::cout << "Time: " << anotherTime << std::endl;
取消引用代码中的指针,如下所示:
MyTime* anotherTime = new MyTime(6, 31, 27, 0, 0);
std::cout << "Time: " << *anotherTime << std::endl;
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