字符串错误，正在保存数据(带有链表)

Question

我正在处理这个链表项目，我应该将员工数据保存在链表中并应用一些操作。我是c++的新手，所以我会给你看我的代码，请帮我。我的问题是：

1-每当我输入全名时，程序就会无限循环！我使用了getline((方法，但仍然不起作用。但如果我只输入第一个名字，效果很好。

2-如果我使用我创建的方法，我不知道如何显示或查找员工信息，它会给我随机数据。我的代码：

main：

#include<iostream>
#include<string>
#include <fstream>
#include "EmployeeList.cpp"
using namespace std;
int main(){
string name;
int choice, id;
double Salary;
EmployeeList* employee = new EmployeeList();
cout<<"***********************************************"<<endl;
cout<<"**********EMPLOYEE MANAGEMENT SYSTEM***********"<<endl;
cout<<"***********************************************nn"<<endl;
do{
cout<<" 1--->Press 1 to INSERT EMPLOYMENT:"<<endl;
cout<<" 2--->Press 2 to FIND EMPLOYMENT INFO ID:"<<endl;
cout<<" 3--->Press 3 to DELETE EMPLOYMENT ID:"<<endl;
cout<<" 4--->Press 4 to SEARCH HIGHEST SALARY:"<<endl;
cout<<" 5--->Press 5 to DISPLAY ALL EMOLYMENTS:"<<endl;
cout<<" 6--->Press 6 to REVERSE DISPLAY ALL EMPLOYMENTS:"<<endl;
cout<<" 7--->Press 7 to EXIT:"<<endl;
cout<<"n Enter Your Choice: ";
cin>>choice;
switch(choice){
case 1:
cout<<endl;
cout<<"Enter name : ";
getline(cin,name);
cout<<"Enter ID : ";
cin>>id;
cout<<"Enter Salary : ";
cin>>Salary;
cout<<endl;
employee ->INSERTEMPLOYMENT(name,id,Salary);
break;
case 2:
cout<<endl;
cout<<"Enter ID : ";
cin>>id;
cout<<endl;
employee->FINDEMPLOYMENTINFOID(id);
break;
case 3:
cout<<endl;
cout<<"Employee ID : ";
cin>>id;
cout<<endl;
employee-> DELETEEMPLOYMENTID(id);
break;
case 4:
cout<<endl;
employee->SEARCHHIGHESTSALARY();
break;
case 5:
cout<<endl;
employee->DISPLAYALLEMPLOYMENTS();
break;
case 6:
cout<<endl;
employee->REVERSEDISPLAYALLEMPLOYMENTS(employee->getHead());
break;
case 7:
exit(0);
default:
cout<<"Invalid choice."<<endl;
break;
}
} while (true);

return 0;
}

员工：

#include<iostream>
#include<stdlib.h>
using namespace std;
class Employee{
private:
string name;
int ID;
double Salary;
Employee* next;
public:
//Constructor
Employee (string n, int id, double s){
n = name;
id = ID;
s = Salary;
this->next = NULL;
}
//Setters
void setName(string n){
n = name;
}
void setID(int id){
id = ID;
}
void setSalary(double s){
s = Salary;
}
void setNext(Employee* next){
this->next = next;
}
//Getters
string getName(){
return name;
}
int getID(){
return ID;
}
double getSalary(){
return Salary;
}   
Employee* getNext(){
return this->next;
}
};

员工列表：

#include "Employee.cpp"
using namespace std;
class EmployeeList{
private:
Employee* head;
public:
//Constructor
EmployeeList(){         
head = NULL;
}
//getter
Employee* getHead(){
return head;
}
//insert method 
void INSERTEMPLOYMENT (string name, int id, double Salary){
Employee* newnode = new Employee(name, id, Salary);
if(head == NULL){
head = newnode;
} 
else {
Employee* temp = head;
while(temp->getNext() != NULL){
temp = temp->getNext();
}
temp->setNext(newnode);
}
cout<<"The employee record was added successfully."<<endl;
cout<<"------------------------------------------"<<endl;
}
//info
void FINDEMPLOYMENTINFOID (int id){
Employee *temp = head;
bool status = false;
while(temp != NULL){
if(temp->getID() == id){
status = true;
cout<<"Employee Name = "<<temp->getName()<<endl;
cout<<"Employee ID = "<<temp->getID()<<endl;
cout<<"Employee Salary = "<<temp->getSalary()<<endl;
cout<<"---------------------------------------"<<endl;
}
temp = temp->getNext();
}
if(status == false) {
cout<<"Employee is not found in the list."<<endl;
cout<<"--------------------------------------------"<<endl;
}
}
//display
void DISPLAYALLEMPLOYMENTS (){
Employee* temp = head;
while(temp != NULL){
cout<<"Employee Name = "<<temp->getName()<<endl;
cout<<"Employee ID = "<<temp->getID()<<endl;
cout<<"Employee Salary = "<<temp->getSalary()<<endl;
cout<<"---------------------------------------"<<endl;
temp = temp->getNext();
}
}
};

Answer 1

这里有两个主要问题。

1. 赋值是不可逆的。。。

   在Employee类中，大多数赋值是颠倒的。构造函数是：

   //Constructor
   Employee (string n, int id, double s){
   n = name;     // replace the local copy of n with name!!!
   id = ID;
   s = Salary;
   this->next = NULL;
   }
   

   它当然应该是：

   //Constructor
   Employee(string n, int id, double s) {
   name = n;
   ID = id;
   Salary= s;
   this->next = NULL;
   }
   

   这就是为什么Employee实例只包含垃圾值的原因。。。

2. 混合getline和流提取器是危险的

   当您使用流提取器读取整数值时，流指针会直接位于最后一个数字之后，因此就在行尾之前。如果在那之后立即使用getline，您将只读取换行符，并将获得一个空字符串。一个简单的技巧是始终跳过空行：

   case 1:
   cout << endl;
   cout << "Enter name : ";
   for (;;) {
   getline(cin, name);
   if (name != "") break;
   }
   ...
   

这应该足以解决眼前的问题。但其他人仍然存在：

• 您永远不会测试输入流的状态：如果用户键入了不正确的数字，您将不会检测到它，并且您的程序将具有错误的行为

• 作为链表的EmployeeList应该表现为一个容器：它应该自动销毁其析构函数中的元素。由于这一点和5的规则，它还应该有一个自定义(或已删除(的复制构造函数和复制赋值运算符。

• 构造函数有一种特殊的语法，可以直接初始化它们的成员，而不是先默认初始化它们，然后分配给它们。它不仅效率稍高，而且是初始化const成员的唯一方法。因此，惯用的方法应该是：

  //Constructor
  Employee (string n, int id, double s)
  : name(n), ID(id), Salary(s), next(nullptr) {}