C ++ LNK2019无法解析的外部符号

Question

我已经阅读了很多有关LNK2019的帖子，但无法解决此错误。

这是我的代码：

Time.h：

#ifndef PROJECT2_TIME_H
#define PROJECT2_TIME_H

#include<iostream>
using std::ostream;

namespace Project2  
{
class Time
{
    friend Time& operator+=(const Time& lhs, const Time& rhs);
    friend ostream& operator<<(ostream& os, const Time& rhs);
public:
    static const unsigned secondsInOneHour = 3600;
    static const unsigned secondsInOneMinute = 60;
    Time(unsigned hours, unsigned minutes, unsigned seconds);
    unsigned getTotalTimeAsSeconds() const;
private:
    unsigned seconds;
};

Time& operator+=(const Time& lhs, const Time& rhs);
ostream& operator<<(ostream& os, const Time& rhs);

}

#endif

Time.cpp：

#include "Time.h"


Project2::Time::Time(unsigned hours, unsigned minutes, unsigned seconds)   
{
    this->seconds = hours*secondsInOneHour + minutes*secondsInOneMinute + seconds;
}

unsigned
Project2::Time::getTotalTimeAsSeconds() const
{
return this->seconds;
}


Project2::Time&
Project2::operator+=(const Time& lhs, const Time& rhs)
{
Time& tempTime(unsigned hours, unsigned minutes, unsigned seconds);
unsigned lhsHours = lhs.seconds / Time::secondsInOneHour;
unsigned lhsMinutes = (lhs.seconds / 60) % 60;
unsigned lhsSeconds = (lhs.seconds / 60 / 60) % 60;
unsigned rhsHours = rhs.seconds / Time::secondsInOneHour;
unsigned rhsMinutes = (rhs.seconds / 60) % 60;
unsigned rhsSeconds = (rhs.seconds / 60 / 60) % 60;
return tempTime(lhsHours + rhsHours, lhsMinutes + rhsMinutes, lhsSeconds + rhsSeconds);
}

ostream&
Project2::operator<<(ostream& os, const Time& rhs)
{
unsigned rhsHours = rhs.seconds / Time::secondsInOneHour;
unsigned rhsMinutes = (rhs.seconds / 60) % 60;
unsigned rhsSeconds = (rhs.seconds / 60 / 60) % 60;
os << rhsHours << "h:" << rhsMinutes << "m:" << rhsSeconds << "s";
return os;
}

main.cpp只是创建Time对象并使用重载的运算符，似乎没有问题（提供了这些代码，所以它们本身也很好）。

我试图删除所有“时间”符号后面的“＆”，但遇到了同样的错误。

这是错误消息：

错误1错误LNK2019：未解决的外部符号“类Project2 :: Time和__cdecl tempTime（unsigned int，unsigned int，unsigned int）”（？tempTime @@ YAAAVTime @ Project2 @@ III @ Z）在函数“类Project2中引用：: Time＆__cdecl Project2 :: operator + =（类Project2 :: Time const＆，类Project2 :: Time const＆）“（?? YProject2 @@ YAAAVTime @ 0 @ ABV10 @ 0 @ Z）c：\\ Users \\ Eon-Gwei \\ documents \\ visual studio 2013 \\ Projects \\ c ++ III_Project2_GW \\ c ++ III_Project2_GW \\ Time.obj c ++ III_Project2_GW

Answer 1

Time& tempTime(unsigned hours, unsigned minutes, unsigned seconds); 声明一个名为tempTime的函数并return tempTime(lhsHours + rhsHours, lhsMinutes + rhsMinutes, lhsSeconds + rhsSeconds); 调用该函数。 由于该函数在任何地方都没有实现，因此会出现链接器错误。

由于应该假定 operator +=返回对其调用对象的引用，因此您应该通过 this修改对象的成员变量，而不是创建新的 Time ，并返回 *this 。 编辑： operator +=任何明智的实现都将修改左侧操作数，而不是创建新对象。 我建议您重新考虑操作员的工作方式。