编译程序时出现错误，并且不确定如何使用适当的函数来解决此问题，以解决此问题。我将在下面发布Main，Header和CPP文件，并附上所需的结果。任何帮助/提示，不胜感激，谢谢！
错误我得到：
Error C2679 binary '+=': no operator found which takes a right-hand operand of type 'double'
更新：
我已经修复了更新“ =”运算符重载的代码，除了1行输出外，它都在工作。
这是我得到的输出：
A：没有名字：$ 0.00
B：节省：$ 10000.99
C：支票：$ 100.99
A：没有名字：$ 10101.98
B：节省：$ 10000.99
C：支票：$ 100.99
A：联合：$ 0.00
B：节省：$ 10000.99
C：支票：$ 100.99
A：节省：$ 10101.98
B：节省：$ 10101.98
C：支票：$ 100.99
A：节省：$ 10302.98
B：节省：$ 10302.98
C：支票：$ 201.00
由于某些原因，“ JOINT”余额变为0，我不确定为什么。它应该显示为$ 10101.98
主要的：
#include <iostream>
#include "Account.h"
using namespace std;
void displayABC(const Account& A,
const Account& B,
const Account& C) {
cout << "A: " << A << endl << "B: " << B << endl
<< "C: " << C << endl << "--------" << endl;
}
int main() {
Account A;
Account B("Saving", 10000.99);
Account C("Checking", 100.99);
displayABC(A, B, C);
A = B + C;
displayABC(A, B, C);
A = "Joint";
displayABC(A, B, C);
A = B += C;
displayABC(A, B, C);
A = B += C += 100.01;
displayABC(A, B, C);
return 0;
}
源文件：
#include <iomanip>
#include <cstring>
#include "Account.h"
using namespace std;
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(bool gotoNewline)const {
cout << (name_[0] ? name_ : "No Name") << ": $" << setprecision(2) << fixed << balance_;
if (gotoNewline) cout << endl;
}
Account& Account::operator+=(Account& other) {
balance_ += other.balance_;
return *this;
}
Account& Account::operator=(const Account& ls) {
strcpy(name_, ls.name_);
balance_ = ls.balance_;
return *this;
}
Account operator+(const Account &one, const Account &two) {
return Account(one.balance_ + two.balance_);
}
ostream& operator<<(ostream& os, const Account& A) {
A.display();
return os;
}
头文件：
#ifndef _ACCOUNT_H__
#define _ACCOUNT_H__
#include <iostream>
class Account {
char name_[41];
double balance_;
public:
Account();
Account(double balance);
Account(const char name[], double balance = 0.0);
void display(bool gotoNewline = true)const;
Account& operator+=(Account& other);
Account& operator=(const Account & other);
friend Account operator+(const Account &one, const Account &two);
};
std::ostream& operator<<(std::ostream& os, const Account& A);
};
#endif
最佳答案
将+=运算符方法更改为：
Account& operator+=(const Account& other);