c++ runtime error with code (classes/constructors) - c++

Question

I seem to be getting two errors:
A) The first iteration of the loop prints out values fine, however if I press 'y' to go for round 2 it autofills the first number as '0/0'. Shown in the image
B) I want to use the third constructor to set numerator = num and denominator = den However, it just seems to set to default values so I made a temporary fix by commenting out the constructor "Rational(num, den)" and physically wrote out numerator = num; and denominator = den;
Any help is appreciated!
// Add appropriate headers
#include <iostream>
#include <cstdlib>
#include <string>
#include <cmath>
#include <sys/time.h>
using namespace std;
/* KEEP THIS COMMENT
* class Rational
* represents a Rational number. Remember rational means ratio-nal
* which means there is a numerator and denominator having
* integer values. Using good ADT techniques, we have made member
* variable private (also known as instance variables) and made member
* functions public.
*/
class Rational
{
private:
int numerator;
int denominator;
public:
// ToDo: Default Constructor
Rational();
// ToDo: Constructor that takes int numerator
Rational(int i);
// ToDo: Constructor that takes int numerator and int denominator
Rational(int p, int q);
// ToDo: Member function to read a rational in the form: n/d
void input();
// ToDo: Member function to write a rational as n/d
void output();
// ToDo: declare an accessor function to get the numerator
int getNumerator();
// ToDo: declare an accessor function to get the denominator
int getDenominator();
// ToDo: delcare a function called Sum that takes two rational objects
// sets the current object to the sum of the given objects using the
// formula: a/b + c/d = ( a*d + b*c)/(b*d)
void sum(Rational a, Rational b);
// test if two rational numbers are equal.
bool isEqual(const Rational& op);
};
int main()
{
// ToDo: declare three rational objects using the default constructor
Rational a, b, c;
char answer='Y';
// Main loop to read in rationals and compute the sum
do {
cout << "\nEnter op1 (in the format of p/q): ";
a.input();
//Debug line
a.output();
// ToDo: use your input member function to read the first rational
cout << "\nEnter op2 (in the format of p/q): ";
// ToDo: use your input member function to read the second rational
b.input();
//Debug line
b.output();
// ToDo: use the third rational to call Sum with first and second as parameters
c.sum(a, b);
cout << "\nThe sum of op1 and op2 is: ";
c.output();
// ToDo: ouptput the third rational
cout << endl;
cout << "\nTry again (Y/N)?";
cin >> answer;
} while (answer == 'y' || answer == 'Y');
// ToDo: test getters
cout << "\nC's numerator is: " << c.getNumerator() << endl;
cout << "\nC's denominator is: " << c.getDenominator() << endl;
// TODO: Use two constructors to declare a whole number 3/1 and a 4/5
// TODO: Use output to print both rationals
//cout << .output() << " " << .output() << endl;
return 0;
}
// ToDO: Implement your class member functions below.
Rational::Rational()
{
numerator = 0;
denominator = 1;
}
Rational::Rational(int i)
{
numerator = i;
denominator = 1;
}
Rational::Rational(int p, int q)
{
numerator = p;
denominator = q;
}
void Rational::sum(Rational a, Rational b)
{
int num = (a.numerator*b.denominator + a.denominator*b.numerator);
int den = (a.denominator*b.denominator);
numerator = num;
denominator = den;
}
void Rational::input()
{
string in;
int num,den;
//cout << "Enter a rational number in the form of x/y : ";
getline(cin, in);
// find the index position of /
int indx = in.find("/");
// seperator out the numerator
num = atoi(in.substr(0, indx).c_str());
// seperate out the denominator
den = atoi(in.substr(indx+1, in.length()).c_str());
// Rational(num, den);
//cout <<num << " " <<den << endl; // Debug Line
numerator = num;
denominator = den;
//cout <<numerator << " " <<denominator << endl; // Debug Line
}
void Rational::output()
{
cout << numerator << "/" << denominator;
}
// Two getter functions
int Rational::getNumerator()
{
return numerator;
}

Answer 1

Here is the solution. Let me know if you don't know what Buffer is or why do we flush it. I'll try to explain further.
cout << "\nTry again (Y/N)?"; //In main
cin >> answer;
cin.ignore(256,'\n'); // Flush the buffer
Here is the explanation.
Yeah, it(cin) will work fine, As cin doesn't have any issue with buffer. But the issue is with getline. Now let's explain why. First, you need to know how the computer takes input in the program. Whenever you cin some value like cin>>x. The value doesn't directly go into x. Firstly it is stored in some temporary location named buffer. That's why you can press backspace on your console. If it directly writes in more variable(memory), you can't press backspace. It means suppose entering string you wrote "appke", but you want to write "apple " you can press backspace(until you haven't pressed enter). Now, what happens you take input with cin as you give input and press Enter for the next input. Like in your case you pressed "y" and then enter(which is denoted by "\n"). So as enter data it goes into the buffer and then you press enter that enter also goes into buffer but system picks only data from buffer like "y" in your case. So your buffer still has "\n" from previous data. Now come to getline. It has one parameter named "delim" [See-here]http://www.cplusplus.com/reference/string/string/getline/ who tells getline when to stop taking input. By default its values is "\n". Now from your previous entry, your buffer has already "\n" in it. So as getline comes in contact with the buffer, it sees "\n", and it thinks that data for input is present as he finds "\n" in the buffer which is the indicator when to stop. That's why it didn't ask for input. Now come to a solution. If ever you think to use getline after cin. All you need to do is to remove "\n" from the buffer. So all you do is cin.ignore("\n"). You are asking cin to ignore "\n" which was present in the buffer. So when control goes to getline. It ignores the already present "\n" in the buffer and works normally.