Program that checks whether numbers are already in an array - c++

Question

This program is supposed to accept values from the keyboard, and require that the user re-enter the value for the employee's id number. However it keeps outputting "Invalid variable" even if I enter a correct value. It needs to only output that if the value is already been entered. For example
if I enter "3453" as the id number it will still output "Invalid Variable" even if I have not entered that number before.
#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;
}
}
}
My program also will not accept a value of more than one letter into the name. If I try and enter that into the program, the program keeps printing "Invalid entry" until I hit ctrl+c.

Answer 1

for (int i = 0; i < SIZE; ++i)
This checks every element in the array, including the one you have just read. You probably meant to put
for (int i = 0; i < count; ++i)
which will check every element up to (but not including) the one you have just read.

Answer 2

in
for (int i = 0; i < SIZE; ++i)
{
if (employee[i].idNum == employee[count].idNum)
error = 1;
}
employee[count] is one of the employee[i] you're going to compare against which means at some point you will
if (employee[count].idNum == employee[count].idNum)
Which is guaranteed to be true.
But if instead you
int tempId;
cin >> tempId;
for (int i = 0; i < SIZE; ++i)
{
if (employee[i].idNum == tempId)
error = 1;
}
and then set
employee[count].idNum = tempId;
at some later time, you can avoid this problem.
Addendum: I recommend picking this logic up and placing it in its own function. That way A) you don't have to repeat it inside the loop a few lines down where you're repeating the check for each retry and it gets the logic out of the way of the rest of the code. B) you can use the same function later for any other "Does this employee exist?" checks you need to write in the future.
In general You want to have lots of simple, easily-tested functions over one monolithic jack-of-all-trades.