如何在C++中的同一函数中使用字符串和双精度

How to use strings and doubles in the same function in C++

本文关键字:字符串 双精度 函数 C++      更新时间:2023-10-16

对于一个类项目,我需要在一个函数中获取客户的全部信息(包括他们的姓名、房间号等(,所以我需要在同一函数中使用doubles和string。有什么办法可以这样做吗?如果没有,还有其他办法吗?

该项目希望我们通过引用传递值。

//Prints a statement for each overnight customer.
#include <iostream>
#include <string>
#include <math.h>
#include <iomanip>
using namespace std;
const double sales_tax_rate = 0.055;
//Function prototypes
string customerInformation(string &customer_name, string &date_of_bill, 
string &hotelormotel_name, double &room_rate, double &number_of_nights, 
double &phone_charges, double &room_number);
int main()
{
string customer_name, dummy, date_of_bill, hotelormotel_name;
double room_rate = 0;
double number_of_nights = 0;
double phone_charges = 0;
double room_number = 0;
double room_cost;
double subtotal;
double total;
double taxes;
int counter = 1;
char repeat;
do
{
counter += 1;
taxes = 0;
total = 0;
subtotal = 0;
room_cost = 0;
customerInformation(customer_name, room_rate, number_of_nights, 
phone_charges, room_number, date_of_bill, hotelormotel_name);
room_cost = room_rate * number_of_nights;
taxes = sales_tax_rate * room_cost;
subtotal = taxes + room_cost;
total = subtotal + phone_charges;
cout << hotelormotel_name << endl << endl;
cout << "Date:             " << date_of_bill << endl;
cout << "Customer's Name:  " << customer_name << endl;
cout << "Room Number:      " << room_number << endl;
cout << "Number of Nights: " << number_of_nights << endl;
cout << setprecision(2) << fixed << endl;
cout << "Room Rate:     $" << room_rate << endl;
cout << "Room Cost:     $" << room_cost << endl;
cout << "Taxes:         $" << taxes << endl;
cout << "Subtotal:      $" << subtotal << endl << endl;
cout << "Phone Charges: $" << phone_charges << endl << endl;
cout << "TOTAL DUE:     $" << total << endl << endl;
cout << "Thank you for staying at " << hotelormotel_name << "!" << 
endl;
cout << "Drive safely and please come again!" << endl << endl;
cout << "Would you like to run this program again? (Y or N): ";
cin >> repeat;
getline(cin, dummy);
cout << endl;
} while (repeat == 'Y' || repeat == 'y');
}
string customerInformation(string &customer_name, string &date_of_bill, 
string &hotelormotel_name, double &room_rate, double &number_of_nights, 
double &phone_charges, double &room_number)
{
cout << "Please enter the following information: " << endl;
cout << "Hotel/Motel Name: ";
getline(cin, hotelormotel_name);
cout << "Customer Name:    ";
getline(cin, customer_name);
cout << "Date:             ";
getline(cin, date_of_bill);
cout << "Room Number:      ";
cin >> room_number;
do
{
cout << "Number of Nights: ";
cin >> number_of_nights;
if (number_of_nights <= 0)
cout << "Error: Invalid data entered, please try again.n";
} while (number_of_nights <= 0);
cout << endl;
do
{
cout << "Room Rate:     $";
cin >> room_rate;
if (room_rate <= 0)
cout << "Error: Invalid data entered, please try again.n";
} while (room_rate <= 0);
do
{
cout << "Phone Charges: $";
cin >> phone_charges;
if (phone_charges < 0)
cout << "Error: Invalid data entered, please try again.n";
} while (phone_charges < 0);
cout << endl;
return date_of_bill, customer_name, room_number, room_rate, 
number_of_nights, phone_charges, hotelormotel_name;
}

为什么不创建一个包含所有信息的结构?

struct customerinfo {
string customer_name;
string date_of_bill;
string hotelormotel_name;
double room_rate;
double number_of_nights; 
double phone_charges; 
double room_number;
};

然后从类型struct返回一个变量。

是的。有一种使用structs的替代方法,您可以按照创建自己的structs

struct customerInfo { 
string customer_name, dummy, date_of_bill, hotelormotel_name;
double room_rate;
double number_of_nights;
double phone_charges;
double room_number;
double room_cost;
double subtotal;
double total;
double taxes;
};

然后可以将结构类型的对象传递给函数,如下所示

//Function prototypes
string customerInformation(customerInfo tempCustomer);

您可以按如下方式使用此tempCustomer

string customerInformation(customerInfo tempCustomer){
cin>>tempCustomer.customer_name;
//you can call all attributes and assign them values accordingly
//to return a particular value you can do
return tempCustomer.customer_name;
}