自编程以来已经很长时间了,我的问题如下。
如果输入的付款率不正确,我希望创建一个返回入口点的循环。我希望它打印出错误通知,如果它不符合标准,否则转到你已经符合标准。我也在字符串Laborer的数据类型上收到错误,因为没有定义。 IntelliSense:标识符“Laborer”未定义。我一直试图解决这个问题,如下所示。任何方向都将不胜感激。
#include "stdafx.h"
#include "stdlib.h"
#include <iostream>
#include <iomanip>
#include <fstream>
#include <cctype>
#include <conio.h>
#include <cstdlib>
#include <cstring>
using namespace std;
int eel;
int hpr;
int kbhit()
void pause()
{
cout << "Press any key to continue....";
while(1)
{
if(kbhit())
{
break;
}
}
}
int _tmain(int argc, _TCHAR* argv[])
{
int eel = Laborer;
int hpr = 0;
while (hpr < 20.00 || hpr >49.99)
{
cout <<"\n\t\t Enter Allowable Pay Rate For The Employee Position."<<endl;
cout <<"\n\n\t\t =====================================================";
cout <<"\n\t\t The Allowable Pay Rate Is: 20.00 to 49.99 per hour."<<endl;
cout <<"\n\n\t\t =====================================================";
cout <<"\n\t\t Enter Correct Pay Rate For The Position of "<<eel <<": ";
cin >> hpr;
cout <<"\n\n\t\t =====================================================";
if (hpr < 20.00 || hpr >49.99){
{
cout<<"\n\n\t\t XX ERROR! ERROR! XX";
cout<<"\n\n\t\t YOU HAVE ENTERED AN INCORRECT PAY RATE FOR THE POSITION";
cout<<"\n\n\t\t PLEASE RE-ENTER THE CORRECT RATE FOR THE POSITION";
}
if (hpr >= 20 && hpr <= 49.99)
{
cout<<"\n\n\t\t =========================================================";
cout<<"\n\t\t You Entered a Correct Pay Rate for the Position of "<< eel <<endl;
cout<<"\n\t\t Employee Hourly Payroll Rate Is: "<<hpr <<endl;
cout<<"\n\n\t\t =========================================================";
}
pause(); //To stop program to see if the loop is correct
}
}
答案 0 :(得分:2)
您不能将未知变量'Laborer'分配给整数。通过你的目标看起来..
我认为您需要做的是将int eel = Laborer;
更改为string eel = "Laborer";