Question

我试图超载std::ostream& operator<<这样做：

#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();
};

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。

我做错了什么？

Answer 1

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;

Answer 2

您的运算符需要const引用的实例对象，但是您给它一个指针。可以std::cout << "Time: " << *anotherTime << std::endl;（注意*）或添加其他运算符：

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

Answer 3

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

在另一个时间实例上面的代码是myTime *，但它必须是myTime。

您有两种解决方法可以解决此问题：

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;

重载运算符std :: ostream＆amp;运营商LT;＆LT;打印实例内存地址

3 个答案: