不确定Unix为什么不喜欢我的代码,无论出于什么原因,这些代码在Visual Studio上都能很好地工作,我需要使用shell客户端进行赋值,当我尝试这样做时,我会收到这些错误。如果这不是最好的问题,我也很抱歉,我只是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
我不熟悉这些错误中的任何一个,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] == ' ')
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;
}
}
最后是主要的:
#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;
}
您的许多问题纯粹是学术性的。
接受Account&
的operator=()
(和其他函数)应该接受const参数:
Account& operator=( const Account& ls );
同样,当您说A = "Joint"
时,您正试图分配一个常量字符串,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=()
仍然应该返回对*this
的引用。
不幸的是,习惯这类错误的最好方法是最大限度地提高VS的错误报告并修复代码,这样就不会出现错误。
希望这能有所帮助。