c++ - 代码在 visual studio 中运行良好，但是当转移到 Unix 时我收到错误？

Question

不确定为什么 Unix 出于某种原因不喜欢我的代码，该代码在 Visual Studio 上完美运行，我需要使用 shell 客户端提交作业，当我尝试这样做时，我收到这些错误。如果这不是最好的问题，我也很抱歉，我真的很困惑为什么这不起作用。以下是错误:

w6_in_lab.cpp: In function âint main()â:
w6_in_lab.cpp:22:11: error: no match for âoperator=â in âA = B.sict::Account::operator+(C)â
w6_in_lab.cpp:22:11: note: candidates are:
Account.h:15:12: note: sict::Account& sict::Account::operator=(sict::Account&)
Account.h:15:12: note:   no known conversion for argument 1 from âsict::Accountâ to âsict::Account&â
Account.h:17:9: note: char* sict::Account::operator=(char*)
Account.h:17:9: note:   no known conversion for argument 1 from âsict::Accountâ to âchar*â
w6_in_lab.cpp:23:7: warning: deprecated conversion from string constant to âchar*â [-Wwrite-strings]
w6_in_lab.cpp:25:10: error: no match for âoperator=â in âA = B.sict::Account::operator+=((* & C))â
w6_in_lab.cpp:25:10: note: candidates are:
Account.h:15:12: note: sict::Account& sict::Account::operator=(sict::Account&)
Account.h:15:12: note:   no known conversion for argument 1 from âsict::Accountâ to âsict::Account&â
Account.h:17:9: note: char* sict::Account::operator=(char*)
Account.h:17:9: note:   no known conversion for argument 1 from âsict::Accountâ to âchar*â

我对这些错误中的任何一个都不熟悉，因为我不太熟悉重载运算符，而且我从不使用 Unix 进行 C++ 赋值。

这是 .h 文件:

#ifndef SICT_ACCOUNT_H__
#define SICT_ACCOUNT_H__
#include <iostream>
#include <cstring>
#include <iomanip>
namespace sict{
    class Account{
        char _name[41];
        double _balance;
    public:
        Account();
        Account(double balance);
        Account(const char name[], double balance);
        Account(const char name[]);
        Account& operator=(Account& ls);
        Account operator+=(Account& ls);
        char* operator=(char* ls);
        void display()const;
        double getBal();
        char* getName();

        friend double operator+=(double& ls, Account& rs);
        Account operator+(Account ls);

    };

    std::ostream& operator<<(std::ostream& ls, Account& rs);
};

#endif

.cpp 文件:

#define _CRT_SECURE_NO_WARNINGS
#include <cstring>
#include <iomanip>
#include "Account.h"
using namespace std;
namespace sict{
    Account::Account(){
        _name[0] = 0;
        _balance = 0;
    }

    Account::Account(double balance){
        _name[0] = 0;
        _balance = balance;
    }

    Account::Account(const char name[], double balance){
        strncpy(_name, name, 40);
        _name[40] = 0;
        _balance = balance;
    }

    void Account::display()const{

        for(int x = 0; x < 40; x++){
            if(_name[x] == '\0')
                x = 40;
            else
                cout << _name[x];
        }

    cout << ": $" << setprecision(2) << fixed << _balance;

    }

    Account Account::operator+(Account ls) {

        return ls._balance + _balance;
    }

    double operator+=(double& ls, Account& rs){
        //cout << ls << endl;
        //cout << rs._balance+ ls << endl;
        return ls+=rs._balance;



        //return rs._balance+=ls;
    }

    Account Account::operator+=(Account& ls){

        return  _balance+=ls._balance;

    }

    Account::Account(const char name[]){
        strncpy(_name, name, 40);
    }

    char* Account::getName(){

        return _name;
    }

    double Account::getBal(){

        return _balance;
    }

    std::ostream& operator<<(std::ostream& ls, Account& rs){
        rs.display();
        return ls;
    }

    Account& Account::operator=(Account& ls){
        if( !strcmp(ls._name,"") &&ls._balance > 0)
        {
            strcpy(_name, "Saving");
        }

        _balance = ls._balance;
        //strcpy(_name, ls._name);

        return *this;

    }

    char* Account::operator=(char* ls){

        strcpy(_name, ls);

        return _name;

    }


}

最后是主要内容:

#include <iostream>
#include <string>
#include "Account.h"
using namespace sict;
using namespace std;

int main(){
  Account A;
  Account B("Saving", 10000.99);
  Account C("Checking", 100.99);
  double value = 0;
cout << A << endl << B << endl << C << endl << "--------" << endl;
  A = B + C;
  A = "Joint";
cout << A << endl << B << endl << C << endl << "--------" << endl;
A = B += C;
cout << A << endl << B << endl << C << endl << "--------" << endl;
value += A;
value += B;
value += C;
cout << "Total balance: " << value << endl;
  return 0;
}

Answer 1

您的许多问题纯粹是学术问题。

采用Account& 的operator=()(和其他函数)应该采用const 参数:

Account& operator=( const Account& ls );

同样，当您说 A = "Joint" 时，您正试图分配一个 const 字符串，但您的运算符只接受一个可变字符串。修复其参数类型:

Account& operator=(const char* ls);

请注意，operator=() 仍应返回对 *this 的引用。

不幸的是，习惯这些错误的最好方法是将 VS 的错误报告调到最大并修复您的代码，这样您就不会出错。

希望这对您有所帮助。