C ++链接错误未解析的函数调用

Question

我现在在线搜索了大约两个小时，我无法弄清楚这段代码有什么问题。我收到了这个错误。

1＆gt; DateAssignment.obj：错误LNK2019：未解析的外部符号＆＃34; public：void __thiscall DayCalculator :: getDate（void）＆＃34; （？getDate @ DayCalculator @@ QAEXXZ）在函数_main中引用 1＆gt; DateAssignment.obj：错误LNK2019：未解析的外部符号＆＃34; public：void __thiscall DayCalculator :: print（void）＆＃34; （？print @ DayCalculator @@ QAEXXZ）在函数_main

中引用

-DayCalculator.h -

#pragma once
using namespace std;

class DayCalculator
{
public:
    void getDate();
    bool checkLeapYear();
    int calcDayOfYear();
    void print();
};

-DateAssignment.cpp -

#include "stdafx.h"
#include "DayCalculator.h"

DayCalculator DayCalc;

int main(int argc, char* argv[])
    {
        DayCalc.getDate();
        DayCalc.print();

        return 0;
    }

-DayCalculator.cpp -

#include "stdafx.h"
#include "DayCalculator.h"
#include <iostream>
#include <string>
#include <sstream>

using namespace std;

string date = "01-01-0001";
int day, month, year;

/*
*   - Get an int input in the form dd-mm-yyyy and break the string into three substrings
*day, month, and year.
*   - Local string 's' is used temporarily to hold substrings of the input as stringstream
*is used to convert them into integers.
*/
void getDate()
{
    string s;

    cout << "Enter the date in the form dd-mm-yyyy(including dashes and zeros): ";
    cin >> date;

    s = date.substr(0,1);
    stringstream(s) >> day;
    s = date.substr(3,4);
    stringstream(s) >> month;
    s = date.substr(6,9);
    stringstream(s) >> year;
}

/*
*   - Check if the specified year is a leap year, if so, February has 29 days as opposed to 28.
*/
bool checkLeapYear()
{
    if(year%4 == 0 && year%100 != 0)
        return true;
    else
        return false;
}

/*
*   - Calculate the number of days in each month of the specified year.
*/
int calcDayOfYear()
{
    int dayOfYear;

    switch(month)
    {
        case 1:
            dayOfYear = day;
            break;
        case 2:
            dayOfYear = day + 31;
            break;
        case 3:
            dayOfYear = day + 59;
            break;
        case 4:
            dayOfYear = day + 90;
            break;
        case 5:
            dayOfYear = day + 120;
            break;
        case 6:
            dayOfYear = day + 151;
            break;
        case 7:
            dayOfYear = day + 181;
            break;
        case 8:
            dayOfYear = day + 212;
            break;
        case 9:
            dayOfYear = day + 243;
            break;
        case 10:
            dayOfYear = day + 273;
            break;
        case 11:
            dayOfYear = day + 304;
            break;
        case 12:
            dayOfYear = day + 334;
            break;
        default:
            dayOfYear = -1;
            break;
    }

    if(dayOfYear < 0 || dayOfYear > 366)
    {
        return -1; 
    }
    else if(dayOfYear > 59 && checkLeapYear() == true)
    {
        if(dayOfYear == 60 && month == 2)
            return 60;
        else
            return dayOfYear+1;
    }
    else
        return day;
}

/*
*   - Output in the form ddd.
*/
void print()
{
    cout << "The day is " + calcDayOfYear();
}