Help finishing code

[CPPRaven]

Is anybody capable of finishing up my code.

If anybody is able to complete this part for me, you will be greatly appreciated.

This is my code. . This is suppose to take in a numerator and denominator, do the math, then reduce it to lowest terms.

So far the code runs fine, but I am not able to finish it.

I need the answer to come out in this form.

The number 2/8 should be reduced to 1/4.

9/6 should be reduced to 3/2, 3/2 should be displayed as 1 1/2.

#include <iostream>
using namespace std;
void showFrac();
void add(int N1, int N2, int D1, int D2, int numrator, int denominator);
void subtract(int N1, int N2, int D1, int D2, int numrator, int denominator);
void multiply(int N1, int N2, int D1, int D2, int numrator, int denominator);
void divide(int N1, int N2, int D1, int D2, int numrator, int denominator);
void Quit();
int main()
{
    showFrac();
    system ("pause");
}
void showFrac()
{
    int N1, N2, D1, D2, numrator = 0, denominator = 0;
    int choose( '\0' );
    while (choose != 5)
    {
        cout<<" \n 1 : Add fraction";
        cout<<" \n 2 : Subtract fraction";
        cout<<" \n 3 : Multiply fraction";
        cout<<" \n 4 : Divide fraction";
        cout<<" \n 5 : Quit";
		cout<<" \n Your choice: ";
        cin >> choose;
        if (choose <=4)
        {
            cout <<"\n Enter first numerator: ";
            cin >> N1;
            cout <<"\n Enter first denominator: ";
            cin >> D1;
            cout <<"\n Enter second numerator: ";
            cin >> N2;
            cout << "\n Enter second denominator: ";
            cin >> D2;
        }
        switch(choose)
        {
        case 1:
            add(N1, N2, D1, D2, numrator, denominator);
            break;
        case 2:
            subtract(N1, N2, D1, D2, numrator, denominator);
            break;
        case 3:
            multiply(N1, N2, D1, D2, numrator, denominator);
            break;
        case 4:
            divide(N1, N2, D1, D2, numrator, denominator);
            break;
        case 5:
            Quit();
			break;
        default:
            cout<<"Enter correct input"<<endl;
        }
    }
}
void add(int N1, int N2, int D1, int D2, int numrator, int denominator)
{
    if(D1 != 0 && D2 != 0)
    {
        numrator = ((N1 * D2) + (N2 * D1));
        denominator = D1 * D2;
        cout << N1 << "/" << D1 << " " << "+" << " " << N2 << "/" << D2 << "=" << numrator << "/" << denominator;
    }
}
void subtract(int N1, int N2, int D1, int D2, int numrator, int denominator)
{
    if(D1 != 0 && D2 != 0)
    {
        numrator = ((N1 * D2) - (N2 * D1));
        denominator = D1 * D2;
        cout << N1 << "/" << D1 << " " << "-" << " " << N2 << "/" << D2 << "=" << numrator << "/" << denominator;
    }
}
void multiply(int N1, int N2, int D1, int D2, int numrator, int denominator)
{
    if(D1 != 0 && D2 != 0)
    {
        numrator = N1 * N2;
        denominator = D1 * D2;
        cout << N1 << "/" << D1 << " " << "*" << " " << N2 << "/" << D2 << "=" << numrator << "/" << denominator;
    }
}
void divide(int N1, int N2, int D1, int D2, int numrator, int denominator)
{
    if(D1 != 0 && D2 != 0)
    {
        numrator = N1 * D2;
        denominator = N2 * D1;
        cout << N1 << "/" << D1 << " " << "/" << " " << N2 << "/" << D2 << "=" << numrator << "/" << denominator;
    }
}
void Quit()
{
    cout << "\n\tThank You\n\n";
}

[ToniAz]

Try this as an alternative, what do you mean but you're not able to finish it?

int main()
{
    int a, b;
    cout << "Enter numerator then denominator:\n";
    cin >> a >> b;

    int d;
    d = 2;
    while (d <= a)
    {
        while (d <= a)
        {
            if (a % d == 0 && b % d == 0)
            {
                a /= d;
                b /= d;
                d = 2;
                continue;
            }
            else d++;
        }
    }

    cout << a << " / " << b << endl;
    return 0;
}

Could you now modify it so it can work on improper fractions too (numerator > denominator) ?

[CPPRaven]

I mean I can't complete what I am trying to do lol which is have the answer come out in a mixed fraction or whole number if possible.

so if the answer is 1/1 then it should display just 1, not 1/1.

and if the answer is 9/6, the it should reduce it to 3/2, and finally display 1 1/2.

[CPPRaven]

Can anybody please help with this ?
I can honestly say that i am stumped, , please help out if you can.

[ToniAz]

so if the answer is 1/1 then it should display just 1, not 1/1.

Can you use an if statement for that, or a switch statement to account for all 3 cases, for example?

[Volatile Pulse]

Here is a code snippet for you to apply to your application about the output of the fraction:

// Check to see if it's a whole number
if (numerator % denominator == 0)
   ;// Display whole number
// Check to see if it's a mixed fraction
else if (numerator / denominator > 0)
   ; // Display mixed fraction
// Anything else would be a simplified fraction
else
   ; // Display simplified fraction 

I didn't look at the variables that you used, but I'm not going to write the entire thing for you either. Hope this helps you get your problem solved.

[cnoeval]

// Check to see if it's a mixed fraction
else if (number / denominator > 0)
   ; // Display mixed fraction 

Shouldn't that be:
else if (numerator / denominator) > 1)

[Volatile Pulse]

Why should it be greater than one? numerator and denominator are ints, do you know how int division works?

[cnoeval]

Duh, brain fart. Integer division discards the fractional part so yes, 0 will work fine. Of course greater than 1 will work too and it seemed a bit more logical for a human reading the code. At least for me anyway...

[Volatile Pulse]

No it wouldn't 3/2 would return 1 which would be false. I understand your logic though. But since C++ does it's division a certain way, it has to be like that.

[helios]

Note that for positive a and b, a/b>1 if and only if a>b.

[Volatile Pulse]

@helios
That only applies to common math, not C++ math. According to your logic, 3 > 2 which satisfies your last criteria, but 3/2 is not > 1, it's equal to 1, which makes your second criteria incorrect, and 3 and 2 are both positive, which means your first is correct. The same is true for any two numbers where:

a > b
a < 2b
a / b = 1

Note: This only applies to non floating point numbers.

[helios]

And your point is...? OP is trying to duplicate the algebraic properties of rational numbers, not of integers.

[Volatile Pulse]

My point is that the OP has the numerator and denominator defined as ints and int division returns an int, therefore he must use C++'s int division to test their answers. That is why your post was irrelevant to hit problem.

Edit: Without having the op change their program and without introducing something they may/may not know or understand, my equation is the correct one.

[helios]

OP has the numerator and denominator defined as ints and int division returns an int, therefore he must use C++'s int division to test their answers

That's a non sequitur.

1. It's possible to predict to some extent the result of an operation without actually carrying it out. For example, a/b>1 if and only if a>b:

inline int abs(int x){
    return (x>=0)?x:-x;
}
//...
if (abs(a)>abs(b)) //... 

2. Less relevantly, there's more than one way to perform integer division.

[Volatile Pulse]

You're suggesting instead of doing the division, simply checking if the two values are greater? That would be right, and I'm sorry for misunderstanding. I thought you were trying to say that the division should be greater than 1. And without casting one of the ints to a floating point, how else can you do int division?