C++ 检查数字是否已在数组中的程序

该程序应该接受来自键盘的值，并要求用户重新输入员工id号的值。然而，即使我输入了正确的值，它也会继续输出“无效变量”。它只需要在已经输入值的情况下输出。例如 如果我输入“3453”作为id号，它仍然会输出“无效变量”，即使我以前没有输入该编号

#include <iostream>

using namespace std;
struct Employee
{
    int idNum;
    double payRate;
    char firstName, lastName;
};

int main()
{
    int error;
    const int SIZE = 5;
    Employee employee[SIZE];
    for (int k = 0; k < SIZE; ++k)
    {
        employee[k].idNum = 0;
        employee[k].payRate = 0;
    }
    for (int count = 0; count < SIZE; ++count)
    {
        error = 0;
        cout << "Enter the employee's id number " << endl;
        cin >> employee[count].idNum;
        for (int i = 0; i < SIZE; ++i)
        {
            if (employee[i].idNum == employee[count].idNum)
                error = 1;
        }
        while (error == 1)
        {
            cout << "Invalid entry. Please enter a new id number " << endl;
            cin >> employee[count].idNum;
            for (int i = 0; i < SIZE; ++i)
            {
                error = 0;
                if (employee[i].idNum == employee[count].idNum)
                    error = 1;

            }
        }
        cout << "Enter the employee's pay rate " << endl;
        cin >> employee[count].payRate;
        cout << "Enter the employee's first name " << endl;
        cin >> employee[count].firstName;
        cout << "Enter the employee's last name " << endl;
        cin >> employee[count].lastName;
        int choice;
        cout << "Enter 1 to search for an employee by id number, enter 2 to                 search by last name, and enter 3 to search by pay "
             << endl;
        cin >> choice;

    }
    int choice;
    cout << "Enter 1 to search for an employee by id number, enter 2 to    search by last name, and enter 3 to search by pay "
         << endl;
    cin >> choice;
    if (choice == 1)
    {
        int idNumC;
        cout << "Enter an id number ";
        cin >> idNumC;
        for (int count = 0; count < SIZE; ++count)
        {
            if (employee[count].idNum == idNumC)
                cout << employee[count].idNum;
        }
    }
    if (choice == 2)
    {
        char name;
        cout << "Enter the employee's last name " << endl;
        cin >> name;
        for (int count = 0; count < SIZE; ++count)
        {
            if (employee[count].lastName == name)
                cout << "ID number: " << employee[count].idNum
                        << " First name: " << employee[count].firstName
                        << " Last Name: " << employee[count].lastName
                        << " Hourly Pay: " << endl;
        }
    }
    if (choice == 3)
    {
        int name;
        cout << "Enter the employee's last name " << endl;
        cin >> name;
        for (int count = 0; count < SIZE; ++count)
        {
            if (employee[count].payRate == name)
                cout << "ID number: " << employee[count].idNum
                        << " First name: " << employee[count].firstName
                        << " Last Name: " << employee[count].lastName
                        << " Hourly Pay: " << endl;
        }
    }
}

#包括
使用名称空间std；
结构雇员
{
int-idNum；
双倍工资率；
char firstName，lastName；
};
int main（）
{
整数误差；
常数int SIZE=5；
雇员[人数]；
对于（int k=0；kidNumC；
对于（int count=0；count
这肯定是真的

但是如果你

int tempId;
cin >> tempId;
for (int i = 0; i < SIZE; ++i)
{
    if (employee[i].idNum == tempId)
        error = 1;
}

稍后，您可以避免此问题

附录：我建议将此逻辑提取出来，并将其放在自己的函数中。这样一来，A）您不必在循环中重复它，每次重试都要在循环的几行中重复检查，它会使逻辑脱离代码的其余部分。B）您可以在以后对任何其他“此员工是否存在”使用相同的函数您将来需要填写的支票

通常，您希望在一个各行各业的单片插孔上拥有许多简单、易于测试的功能。
for（int i=0；i for (int i = 0; i < SIZE; ++i)

这将检查数组中的每个元素，包括刚刚读取的元素

 for (int i = 0; i < count; ++i)

for（int i=0；i

---

将检查每个元素（但不包括）你刚刚读过的那一个。

备份你的文件，然后删除除了输入和验证代码以外的所有内容。然后用调试器或纸上的笔来完成。看看你是否能在这里的人得到答案之前弄清楚。如果你写了一个构造函数，你不需要在外部设置

idNum

或

payRate

。当您检查是否已输入id时，您会检查所有数组元素，包括您刚输入的元素，以便始终找到它。可能会跳过当前项？顺便说一句，如果读取了第一个员工id，数组中有多少个？数组中总共有5个

employee[count].idNum = tempId;

 for (int i = 0; i < SIZE; ++i)

 for (int i = 0; i < count; ++i)