6

我一生都无法弄清楚为什么会出现这些错误，尤其是因为我包含了警卫。

这些是我的错误（请忽略我命名我的计算机的内容）：

1>main.obj : 错误 LNK2005: "class std::basic_ostream > >& __cdecl operator<<(class std::basic_ostream > &,class >Date &)" (??6@YAAAV?$basic_ostream@DU?$ char_traits@D@std@@@std@@AAV01@AAVDate@@@Z) 已经>在loan.obj中定义

1>main.obj : 错误 LNK2005: "class std::basic_ostream > >& __cdecl operator<<(class std::basic_ostream > &,class >Loan &)" (??6@YAAAV?$basic_ostream@DU?$ char_traits@D@std@@@std@@AAV01@AAVLoan@@@Z) 已经>在loan.obj中定义

1>C:\Users\SweetAssSarah\Documents\Visual Studio >2012\Projects\ConsoleApplication4\Debug\a1.exe：致命错误 LNK1169：找到一个或多个乘法 > 定义的符号

这是我的 4 个文件： Main.cpp：

#ifndef _main_cpp
#define _main_cpp

#include<iostream>
#include "loan.h"
#include "date.h"

using namespace std;

void main(){
    const int MAX_SIZE = 80;
    char response[MAX_SIZE];

    Loan sarah("Sarah", "123 Awesomeville ", Date (01,February,2010));

    cout << sarah.getName() << " address: " << sarah.getAddress() << endl;
    cout << "Date: " << sarah.getDate() << endl;

    //keep console open until user types a key and enter
    cout <<"\n\n" << "Press ENTER to continue";
    cin.getline(response, MAX_SIZE);

    return;
}    
#endif

贷款.cpp：

#ifndef _loan_cpp
#define _loan_cpp

#include <iostream>
#include "loan.h"
#include "date.h"

using namespace std;

Loan::Loan(char * aName, char * anAddress, Date aDate){
    name = aName;
    address = anAddress;
    date = aDate;
    cout <<"CONSTRUCTING: " << name << "\n"; 
}    
Loan::~Loan(){
    cout << "DESTRUCTING: " << name << endl;
} 
char * Loan::getName() {return  name;} 
char * Loan::getAddress(){return address;} 
Date Loan::getDate(){return date;}

void Loan:: printOn(ostream & ostr) {
    cout << name << " address: " << address << endl;
}    
#endif

贷款.h：

#ifndef _loan_h
#define _loan_h

#include <math.h> //for the pow() function to do exponentiation
#include <iostream>
#include "date.h"

using namespace std;

class Loan{
public:
    Loan(char *, char *, Date );//constructor
    ~Loan();

    char * getName();
    char * getAddress();
    Date getDate();

    void printOn(ostream & ostr);
private: 
    char * name; 
    char * address;
    Date date; //requires class Date to have a default constructor

};
ostream & operator<<(ostream & ostr, Loan & aLoan) { 
    aLoan.printOn(ostr); 
    return ostr; 
}
#endif

日期.h：

#ifndef _date_h
#define _date_h
#include <iostream>

enum Month {January=1, February, March, April, May, June, July, August,
September, October, November, December};

using namespace std;

class Date{
public:
//  Date() {};
    Date(int aDay = 1, Month aMonth = May, int aYear = 2005){
        day = aDay;
        month = aMonth;
        year = aYear;
}    
void printOn(ostream & o){
    o << day << "/" << month << "/" << year;
}
private:
    int day;
    Month month;
    int year;
};    
ostream & operator<<(ostream & ostr, Date & d) { 
    d.printOn(ostr); 
    return ostr; 
} 
#endif

请帮忙！

4

1 回答 1

16

如评论中所述，<< 运算符需要在 cpp 文件中内联或定义。当您在头文件中定义函数时，它将在包含头文件的每个 cpp 文件中编译。如果在多个 cpp 文件中包含标头，则会出现编译器错误，因为相同的代码将被编译成多个 .obj 文件。链接器不知道要使用哪个 .obj 文件并引发错误。

解决方案 1 - 拆分为 .h 和 .cpp

贷款.h

ostream & operator<<(ostream & ostr, Loan & aLoan);

贷款.cpp

ostream & operator<<(ostream & ostr, Loan & aLoan) { 
    aLoan.printOn(ostr); 
    return ostr; 
}

解决方案 2 - 内联

贷款.h

inline ostream & operator<<(ostream & ostr, Loan & aLoan) { 
    aLoan.printOn(ostr); 
    return ostr; 
}

第二种解决方案将导致编译器通过在代码中发生调用的每个位置内联函数代码来解析函数调用。这会导致编译代码的冗余，对于大型函数应该避免。

解决方案 3 - 静态

贷款.h

static ostream & operator<<(ostream & ostr, Loan & aLoan) { 
    aLoan.printOn(ostr); 
    return ostr; 
}

通过将函数声明为静态，它具有内部链接。您仍然将编译器是否内联它的决定留给编译器。

于 2013-10-04T10:46:54.180 回答