Trapping Bad Entries in a C++ Mortgage calculator program

[promiscuoustx]

The problem is that my code below used to run wonderfully, until the instructor decided that he wants to use characters instead of integers, and wants my code to trap the bad and have the program state that the user did not imput the correct values. Can someone help me figure out how to make my code work that way??? I have the majority of the code done.....but....I still need that last step!! Thanks!!

#include <iostream>        
#include <iomanip>
#include <cmath>
#include <cctype>        
 
 
using namespace std;
int main()
 
{
     double a; //This is the amount of the mortgage the user must enter
     double i; //This is the interest rate the user must enter
     int y; //This is the amount of years for the mortgage the user must enter            
     double mPayment; //This is a variable for ouputting the payment            
     char YesNo = 'Y';
 
 
do //In order to allow the user to be able to re-enter date, we must have a loop
 
{
    cout << "What is the amount of the mortgage?  For example 200000" << endl;
    cout << "Press enter." << endl;
    cin    >>    a;    
    cout << endl;
    cout << "What is the amount of years the mortgage will be financed? For example 30" << endl;
    cout << "Press enter." << endl;
    cin >>    y;
    cout << endl;
    cout << "What is the interest rate? For example 5.75" << endl;
    cout << "Press enter." << endl;
    cin >>    i;
    cout << endl;
 
//These are the variables required to calculate the information the user inputs
double monInterest = i / 12 / 100; //Calcualtes the interest monthly        
int t = y * 12; //This is the loan term in the amount of months
 
//This is the actual formula for calculating the mortgage payment amount 
mPayment = (a * monInterest) / (1-pow((1+monInterest),-t));
 
//This allows the user to view what they entered and also what the monthly payment would be
    cout << "Amount of mortgage = $" << a << endl;
    cout << "Year financed = " << i << "%" << endl;
    cout << "Interest Rate = " << y << " years" << endl;
    cout << endl;
    cout << "Monthly Payment Amount = $" << mPayment << endl;
    cout << endl;
 
 
//This allows the user to either enter in new information, or exit the program
    cout << "If you would like to enter different information, please press Yes." << endl;
    cout << "If you would like to exit this program, please press No." << endl;
    cout << endl;
    cin >> YesNo;
}
 
while ((YesNo == 'Y') || (YesNo == 'y')); //End of the loop    
 
    return 0;
 
}
 

[Banfa]

It's not entirely clear what you mean, are you talking about making the variables y and t type char instead of type int?

While that shouldn't be too much of a problem for y (unlikely to get terms > 255 years) it may be for t which is y * 12.

To ctach the bad case you need to see if y * 12 will be greater than 255

y * 12 > 255

That is actually not possible without breaking the constraints of a char but you can manipulate the equation so that a valid check is made.

Just divide both sides of the equation by 12 (simple algabra)

y * 12 / 12 > 255 / 12

simplying

y > 21.25

since it's integer arithmatic drop the fraction

y > 21

is the error condition

[promiscuoustx]

Ok, maybe I was not making myself clear. Attached is exactly what my instructor said....that might help....

"Modify the mortgage program to input the amount of the mortgage, the term of the mortgage, and the interest rate of the mortgage. Display the mortgage payment amount. Allow the user to loop back and enter new data or quit. Insert comments in the program to document the program. This program should still be a procedural C++ program. The first version of the code is used to allow students to work on the process of writing the code, getting it reviewed, and applying feedback. In the subsequent versions, we add features to the initial version. While accepting input may be necessary to perform a calculation, the emphasis is on error checking the user input. This is not a trivial task.. Also, error checking will require the use of looping and arrays, which we recently addressed and will continue this week.

btw. the values I use to test the input amount are : 123.45 123,45 1a2b.a4 123.4a and a234.56
Your code should identify the bad input, notify the user, and take appropriate action."

I hope this explains better what I am trying to do, but am stuck with. I was thinking I could use isdigit(), but I think that is only used for single integers.

[promiscuoustx]

I have being trying to figure out how I can validate my code using isdigit(), but I am not sure if I can use isdigit() when there is more than one number. I was also told to try islanum(), but I have never heard of this and unsure what it is used for.

[promiscuoustx]

This is the code I have so far.....any help would be most appreciated!!!

#include <iostream>        
#include <iomanip>
#include <cmath>
#include <cctype>        
 
 
using namespace std;
int main()
 
{
     double a; //This is the amount of the mortgage the user must enter
     double i; //This is the interest rate the user must enter
     int y; //This is the amount of years for the mortgage the user must enter            
     double mPayment; //This is a variable for ouputting the payment            
     char YesNo = 'Y';
     int b;
 
do //In order to allow the user to be able to re-enter date, we must have a loop
 
{
    cout << "What is the amount of the mortgage?  For example 200000" << endl;
    cout << "Press enter." << endl;
    cin    >>    a ,b;
    if(isdigit(b))
    {
    cout << "What is the amount of years the mortgage will be financed? For example 30" << endl;
    cout << "Press enter." << endl;
    }
    else
    {
    cout<<"You have not entered a valid entry! Please try again.";
    }
}while ((YesNo == 'Y') || (YesNo == 'y'));   
    cout << endl;
    cout << "What is the amount of years the mortgage will be financed? For example 30" << endl;
    cout << "Press enter." << endl;
    cin >>    y;
    if(isdigit(b))
    {
    cout << endl;
    cout << "What is the interest rate? For example 5.75" << endl;
    cout << "Press enter." << endl;
    cin >>    i;
    cout << endl;
 
//These are the variables required to calculate the information the user inputs
double monInterest = i / 12 / 100; //Calcualtes the interest monthly        
int t = y * 12; //This is the loan term in the amount of months
 
//This is the actual formula for calculating the mortgage payment amount 
mPayment = (a * monInterest) / (1-pow((1+monInterest),-t));
 
//This allows the user to view what they entered and also what the monthly payment would be
    cout << "Amount of mortgage = $" << a << endl;
    cout << "Year financed = " << i << "%" << endl;
    cout << "Interest Rate = " << y << " years" << endl;
    cout << endl;
    cout << "Monthly Payment Amount = $" << mPayment << endl;
    cout << endl;
 
 
//This allows the user to either enter in new information, or exit the program
    cout << "If you would like to enter different information, please press Yes." << endl;
    cout << "If you would like to exit this program, please press No." << endl;
    cout << endl;
    cin >> YesNo;
}
 
while ((YesNo == 'Y') || (YesNo == 'y')); //End of the loop    
 
    return 0;
 
}