need help with fractions in C++

I need to make fractions to be in the simplest form. how to do it .i know i need to call and pass fraction to gcd and simplify but i dont now how to do it so it will work..

Code:

---


int divide()


{


int one4,two4,three4,four4;
int e4,f4;


cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
cin>>one4;
cin>>two4;
cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
cin>>three4;
cin>>four4;


cout<<"Input instruction/_" << endl;


e4 = one4*four4;
f4 = two4*three4;




cout<<"("<< one4 << "/"<< two4 <<") / ("<< three4 <<"/"<< four4 <<") = " << e4 << "/" << f4;


cout << "\n";
return 0;
}





void simplify(int x2 , int y2 ,int gcd )




 {


int a;
int b;






cout<< "Input the nominator and denomiantor of a fraction separated by spaces :" << x2 <<"  "<< y2 << endl;
 
a= x2/gcd;
b= y2/gcd;


cout<<"This fraction equals ("<< a <<"/"<< b << ")"<< endl;








}






int gcd(int x2, int y2)
{
    int i;
    int gcd2;


    for (i=1; i<=x2; i++)
    {


        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
   
            return gcd2;


   
   
}




void input()
{
    int x,y,k;
    cout << "Enter first integer: ";
    cin >> x;
    cout << "Enter second integer: ";
    cin >> y;
    k=gcd(x,y) ;


    cout <<"GCD is "<< k << endl;




    simplify( x , y , k); //calls


}

---

First, fix the layout of your code. Far too many empty lines and no indentation; it's completely unreadable.

Second, state exactly what you want to happen and what actually happens with your code.

i want the fraction result in multiply to be simpler.thats why i was trying to send result from multiply to gcd and after to simplify but im doing something wrong

As CornedBee said:

Quote:

---

First, fix the layout of your code.

---

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;


int multiply()


{
int one3,two3,three3,four3;
int e3,f3;


cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
cin>>one3;
cin>>two3;
cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
cin>>three3;
cin>>four3;
cout<<"Input instruction*_" << endl;
e3 = one3*three3;
f3 = two3*four3;


cout<<"("<< one3 << "/"<< two3 <<") * ("<< three3 <<"/"<< four3 <<") = " << e3 << "/" << f3;
cout << "\n";
return 0;
}
int gcd(int x2, int y2)
{
int i;
int gcd2;
for (i=1; i<=x2; i++)
{
if((x2%i==0) && (y2%i==0))
gcd2 = i;
}
return gcd2;
               
}

void simplify(int x2 , int y2 ,int gcd )
{
int a;
int b;
a= x2/gcd;
b= y2/gcd;
}



void main()


{
        multiply();


        system ("PAUSE");
}

---

You seem to have some extra formatting (e.g., bolding) in the code you posted. And very little indentation. Properly formatted code looks more like this:

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;

int multiply()  /// Should be void multiply() if it only ever returns 0.
{
    /// Why do your variables all end with the number 3?
    int one3,two3,three3,four3;
    int e3,f3;

    /// It's "numerator" not "nominator".
    cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
    cin>>one3;
    cin>>two3;

    cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
    cin>>three3;
    cin>>four3;

    cout<<"Input instruction*_" << endl;
    e3 = one3*three3;
    f3 = two3*four3;

    cout<<"("<< one3 << "/"<< two3 <<") * ("<< three3 <<"/"<< four3 <<") = " << e3 << "/" << f3;
    cout << "\n";
    return 0;
}

/// This doesn't seem to be a proper gcd implementation.
/// Didn't you look at the pseudocode I gave in your original thread?
int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;       
}

/// This function doesn't actually accomplish anything. (And is never called.)
void simplify(int x2 , int y2 ,int gcd )
{
    int a;
    int b;
    a= x2/gcd;
    b= y2/gcd;
}

void main()            /// should be int main() and return 0 at the end
{
        multiply();
        system ("PAUSE");
}

---

Code:

---


int multiply()


{
int one3,two3,three3,four3;
int e3,f3;


cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
cin>>one3;
cin>>two3;
cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
cin>>three3;
cin>>four3;
cout<<"Input instruction*_" << endl;


e3 = one3*three3;
f3 = two3*four3;
gcd(e3,f3); //call
cout<<"("<< one3 << "/"<< two3 <<") * ("<< three3 <<"/"<< four3 <<") = " << e3 << "/" << f3<<this  e3 and f3 i want to be simpler
cout << "\n";
return 0;
}


void simplify(int x2 , int y2 ,int gcd )


 {


int a;
int b;
a= x2/gcd;
b= y2/gcd;
return a,b;




}


int gcd(int x2, int y2)
{
int i;
int gcd2;
for (i=1; i<=x2; i++)
{
if((x2%i==0) && (y2%i==0))
gcd2 = i;
}
return gcd2;


       
       
}


void main()


{
        multiply();


        system ("PAUSE");
}

---

its only part of my assigment ..thats why u can see 3 in the end of letters.. my assigment was to divide multiply add substract fractions and after make them simpler... so forexample 5/10 will be 1/2 .... thats why first i did multiply after i want the integers to move to gcd and simplify and come back simpler to multiply in cout...

in simplify i cant move a,b to my cout in multiply... please help me Im so confussed..

Code:

---

return a,b;

---

You can't return two values from a function like that. Look into passing-by-reference.

im going crazy here

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;


int divide()


{
int one4,two4,three4,four4;
int e4,f4;
int x;


cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
cin>>one4;
cin>>two4;
cout<<"Input the nominator and denominator of a fraction separated by spaces :"<<endl;
cout<<"Input instruction/_" << endl;




e4 = one4*four4;
f4 = two4*three4;


x=gcd(e4,f4);
simplify(e4,f4,x);
cout<<x<<endl;
cout<<"("<< one4 << "/"<< two4 <<") / ("<< three4 <<"/"<< four4 <<") = " << e4 << "/" << f4;
cout << "\n";
return 0;
}


int multiply()


{


int one3,two3,three3,four3;
int e3,f3,p;


cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
cin>>one3;
cin>>two3;
cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
cin>>three3;
cin>>four3;
cout<<"Input instruction*_" << endl;
e3 = one3*three3;
f3 = two3*four3;
p=gcd(e3,f3);  here is telling me(1>fdgfd.cpp.cpp(24): error C3861: 'gcd': identifier not found
1>fdgfd.cpp.cpp(25): error C3861: 'simplify': identifier not found
1>fdgfd.cpp.cpp(48): error C3861: 'gcd': identifier not found
1>fdgfd.cpp.cpp(49): error C3861: 'simplify': identifier not found
1>
simplify(e3,f3,p);


cout<<"("<< one3 << "/"<< two3 <<") * ("<< three3 <<"/"<< four3 <<") = " << e3 << "/" << f3;


cout << "\n";
return 0;
}
void simplify(int x2 , int y2 ,int gcd )
 {


int a;
int b;
a= x2/gcd;
b= y2/gcd;
return (a,b);


}
int gcd(int x2, int y2)
{
    int i;
    int gcd2;


    for (i=1; i<=x2; i++)
    {


        if((x2%i==0) && (y2%i==0))
gcd2 = i;
    }
return gcd2;
}
void main()


{
multiply();
divide();


system ("PAUSE");
}

---

MAybe now somebody can help me ....

Lookup function prototypes
Function prototype - Wikipedia, the free encyclopedia

The above is a guess since your code is not readable.

Tim S.

You've been told to indent your code. You've even been shows how it should look like.
You've even been told main should return int.

Yet, I still do not see you have fixed any of this. Why?
If you expect any more help, then fix what you've already been told to fix!

if i will know how to do it I will be not on this forum.... Yes people told me that my code is unreadable but in my class everybody is writing like that.. so im not catching u guys.. im new to it .. all of u know it ...
my gcd function works
simplify tooo
only thing im asking is for somebody to write me how to connect it together.

Quote:

---

Originally Posted by Agnes Roguska

if i will know how to do it I will be not on this forum.... Yes people told me that my code is unreadable but in my class everybody is writing like that.. so im not catching u guys.. im new to it .. all of u know it ...

---

Don't blame it on others. Because someone else is doing something is not an excuse for you to do it.
We do not expect you to know everything here. But we do expect you to put an effort into learning what you're told, not ignoring it.
If you don't understand how to do something, google it, or if you can't find any good resources that help your understanding, ask for help about it.
We will help you achieve the understanding as long as you are willing. But do not ignore advice or comments if you don't understand them.

Quote:

---

only thing im asking is for somebody to write me how to connect it together.

---

First things first. Fix what you've been told.
If you don't understand something, ask about it.
How difficult is it to change the return type of main?
What is it that is so difficult about indenting?
At least you can become better than your class mates who all write unreadable code and who will not get a programming or will be a pain in the ass to work with.

Quote:

---

Originally Posted by Agnes Roguska

Yes people told me that my code is unreadable but in my class everybody is writing like that.. so im not catching u guys.. im new to it .. all of u know it ...

---

oogabooga gave you an example in post #6. The basic idea is simple: when your code is in a block of scope (e.g., within the body of a function, an if statement, or a loop), you indent by one level. If within the block there is an inner block, you indent again by one level. After the block ends, you decrease the indentation by one level. An indentation can be either a few spaces (my editor is set to produce 4 spaces when I press the tab) or a tab character.

The other thing is blank lines: as a rule of thumb, you should not have more than two consecutive blank lines in your code. Usually, a single blank line to denote some logical separation of the lines of code will suffice.

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;
 
int multiply()
{
    int one3,two3,three3,four3;
    int e3,f3, x;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>one3;
    cin>>two3;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>three3;
    cin>>four3;
 
    cout<<"Input instruction*_" << endl;
    e3 = one3*three3;
    f3 = two3*four3;
        x=gcd(e3,f3);
        simplify(e3,f3,x);
 
    cout<<"("<< one3 << "/"<< two3 <<") * ("<< three3 <<"/"<< four3 <<") = " << e3 << "/" << f3;
    cout << "\n";
   
}
int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
 
int simplify(int x2 , int y2 ,int gcd )
{
    int a;
    int b;
    a= x2/gcd;
    b= y2/gcd;
        return (a,b);
}
 
int main()         
{
    multiply();
    system ("PAUSE");
}

---

gcd is working

Very good. Now you are light years ahead of your class mates.
Now you need to learn how to use references in order to pass back two values:

Code:

---

int simplify(int x2 , int y2 ,int gcd, int& a, int& b)
{
    a= x2/gcd;
    b= y2/gcd;
    return (a,b);
}

Call as
int a, b;
simplify(x2, t2, gcd, a, b);
// a and b will have the results

---

Quote:

---

Originally Posted by Agnes Roguska

gcd is working

---

That's good, though I note that there's a better algorithm (search the Web for Euclid's algorithm).

So, what else do you need help with?

Here's a tip: use descriptive variable names. Instead of one3, two3, three3, and four3, it would be better to name them numerator1, denominator1, numerator2, and denominator2.

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;
 
int multiply()
{
    int numerator,denominator,numerator2,denominator2;
    int x2,t2, x;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator;
    cin>>denominator;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator2;
    cin>>denominator2;
 
    cout<<"Input instruction*_" << endl;
    x2 = numerator*numerator2;
    t2 =denominator*denominator2;
int a, b;
        x=gcd(x2,t2);
        simplify(x2, t2,x, a, b);
       
 
    cout<<"("<< numerator << "/"<<denominator <<") * ("<< numerator2 <<"/"<< denominator2 <<") = " <<a << "/" << b;
    cout << "\n";
   
}
int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
 
int simplify(int x2 , int y2 ,int gcd , int a , int b )
{
   
    a= x2/gcd;
    b= y2/gcd;
        return (a,b);


}
 
int main()         
{
    multiply();
    system ("PAUSE");
}

---

why is showing always that 1>fdgfd.cpp.cpp(22): error C3861: 'gcd': identifier not found
1>fdgfd.cpp.cpp(23): error C3861: 'simplify': identifier not found

Learn function prototypes.

And the pass by reference that's been repeatedly pointed out to you. You seem to struggle to heed the advice given to you.

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;


int simplify(int x2 , int y2 ,int gcd , int a , int b )
{
   
    a= x2/gcd;
    b= y2/gcd;
        return (a,b);
}


 int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
int multiply()
{
    int numerator,denominator,numerator2,denominator2;
    int x2,t2, x;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator;
    cin>>denominator;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator2;
    cin>>denominator2;
 
    cout<<"Input instruction*_" << endl;
    x2 = numerator*numerator2;
    t2 =denominator*denominator2;
    int a, b;
        x=gcd( x2, t2);
        simplify( x2, t2, x, a, b);
       
 
    cout<<"("<< numerator << "/"<<denominator <<") * ("<< numerator2 <<"/"<< denominator2 <<") = " <<a << "/" << b;
    cout << "\n";
    return 0;
}


 


int main()         
{
    multiply();
    system ("PAUSE");
}

---

hmm the result is 0 hahah i know u are giving me tips but im so useless

Read up on references again.

pass by the references - can u give me an easy example .. cos im searching on google and its showing me hard examples.
sorry guys but really programming is too hard for me ..cos im mixing everything

I already did a while ago.

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;
 
 
int simplify(int *x2 , int *y2 ,int *gcd , int *a , int *b )
{
   
    *a= *x2/*gcd;
    *b= *y2/*gcd;
    return (*a,*b);
}
 
 
 int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
int multiply()
{
    int numerator,denominator,numerator2,denominator2;
    int x2,t2, x;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator;
    cin>>denominator;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator2;
    cin>>denominator2;
 
    cout<<"Input instruction*_" << endl;
    x2 = numerator*numerator2;
    t2 =denominator*denominator2;
    int a, b;
    x=gcd( x2, t2);
    simplify( x2, t2, x, a, b);
   
 
    cout<<"("<< numerator << "/"<<denominator <<") * ("<< numerator2 <<"/"<< denominator2 <<") = " <<a << "/" << b;
    cout << "\n";
    return 0;
}
 
 
 
 
 
int main()         
{
    multiply();
    system ("PAUSE");
}

---

can somebody do it for me !!!so i will not disturb u anymore .

There is absolutely no need for pointers. You are only complicating this more than it needs to be.
Go back to your previous code, then go back look at my example on how to use references. If needs be, start a new project and try to get it working, then return to your current project once you got it working.
And no - no one will do it for you! You are the one who needs to learn, not us!

Code:

---



#include "stdafx.h"
#include <iostream>
using namespace std;
 
 
int simplify(int x2 ,int y2 ,int gcd ,int a ,int b )
{
   
    a= x2/gcd;
    b= y2/gcd;


    return simplify;
}
 
 
 int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
int multiply()
{
    int numerator,denominator,numerator2,denominator2 ;
    int x;
        int x2, t2;
        int c;
        int d;


    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator;
    cin>>denominator;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator2;
    cin>>denominator2;
 
    cout<<"Input instruction*_" << endl;
    x2 = numerator*numerator2;
    t2 =denominator*denominator2;
   
    x=gcd( x2, t2);
    simplify(  x2 ,  t2 ,x , c ,  d);
   
 
    cout<<"("<< numerator << "/"<<denominator <<") * ("<< numerator2 <<"/"<< denominator2 <<") = " << c << "/" << d;
    cout << "\n";
    return 0;
}
 
 
 
int main()         
{
    multiply();
    system ("PAUSE");
}

---

far too many uninitialised variables, set things to 0 as often as you can, at least then you have a known value to work with. and at least as a basic protection against bad user input in this case., this thread is littered with examples of you just not taking any notice of good advice given.

and ay caramba :

Code:

---

int simplify(int x2 ,int y2 ,int gcd ,int a ,int b )
{
   
    a= x2/gcd;
    b= y2/gcd;


    return simplify;
}

---

What on earth do you think simplify will be?

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;
 
 
int simplify(int x2 ,int y2 ,int gcd ,int a ,int b )
{
   
    a= x2/gcd;
    b= y2/gcd;


    return simplify( x2, y2, gcd, a,b);
}
 
 
 int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
int multiply()
{
    int numerator,denominator,numerator2,denominator2 ;
    int x;
        int x2, t2;
        int c;
        int d;


    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator;
    cin>>denominator;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator2;
    cin>>denominator2;
 
    cout<<"Input instruction*_" << endl;
    x2 = numerator*numerator2;
    t2 =denominator*denominator2;
   
    x=gcd( x2, t2);
    simplify(  x2 ,  t2 ,x , c ,  d);
   
 
    cout<<"("<< numerator << "/"<<denominator <<") * ("<< numerator2 <<"/"<< denominator2 <<") = " << c << "/" << d;
    cout << "\n";
    return 0;
}
 
 
 
int main()         
{
    multiply();
    system ("PAUSE");
}

---

right erm, yea.

If you want good help, its a two way thing, are you really so unbelievably lazy and rude that you think its ok to just dump wedges of code by way of a response every time? And then expect to get expert help from the board members, when its a bloomin sunday too.

Why not add some words too?????

At this point, you are just guessing.
Work with us - what is it that you fail to understand?

rogster Im not rude... I sitting on this project and really have enough . im dont want to be a programmer its only one of my subjects which i need to pass.. I dont even like it. u think for me its easy to ask like an idiot 10 times this same.. ..
i will be never a programmer..
i really need help to finish it..

Well, even more than you not wanting to be a programmer, we don't want to be robots :)

So, when you post your code, tell us: what's wrong?

Regardless, the subject is there for a reason. Therefore, you need to make sure you acquire the abilities needed to pass the course. It doesn't matter whether you like it or not.
Tell us what you are having problems with - in details, or just forget about getting any help. We cannot help you if you do not work with us, and we will not do your homework for you.

why return simplify( x2, y2, gcd, a,b);
didnt return the values to multiply

What do you expect it to do? That's a function call.

well its a bit of a shaky one, but an oft quoted:

"To ask may be but a moment's shame, not to ask and remain ignorant is a lifelong shame"

We don't mind how often you ask questions, that is good, and it is true that by teaching you learn, so like i say it is a two way thing, I would not bother answering at all otherwise.

You just have to consider how it looks when code is posted without any other message, It comes across very badly, no one is paid here remember.
Fair enough you dont like programming, but as Elysia points out it is a pre requisite of your course, I hated half the crap i had to learn too,( my study was nothing to do with IT or technical by the way) but it is a case of shifting your mindset and concentrating on the long term goal.

To be honest, the problem you are facing is that you are not reading.
You want to return two values from a function. We say: "use references." I even gave you an example in post #18: http://cboard.cprogramming.com/cplus...ml#post1090046
But nowhere in your code do I see you using references.

Code:

---

#include "stdafx.h"
#include <iostream>
using namespace std;
 

int simplify(int x2 ,int y2 ,int gcd ,int a ,int b )
{
   
    a= x2/gcd;(simplify its counting a and b)
    b= y2/gcd;


    return (x2,y2,gcd,a,b); now its returning the x2,y2,gcd,a,b but the value of and b its different
}
 
 
 int gcd(int x2, int y2)
{
    int i;
    int gcd2;
    for (i=1; i<=x2; i++)
    {
        if((x2%i==0) && (y2%i==0))
            gcd2 = i;
    }
    return gcd2;   
}
int multiply()
{
    int numerator,denominator,numerator2,denominator2 ;
    int x;
    int x2, t2;
    int c;
    int d;


    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator;
    cin>>denominator;
 
    cout<<"Input the numerator and denominator of a fraction separated by spaces :"<<endl;
    cin>>numerator2;
    cin>>denominator2;
 
    cout<<"Input instruction*_" << endl;
    x2 = numerator*numerator2;
    t2 =denominator*denominator2;
   
    x=gcd( x2, t2);
    simplify(  x2 ,  t2 ,x , c ,  d); here im sending arguments  to simplify(1)                3)its returning the values here 
 
    cout<<"("<< numerator << "/"<<denominator <<") * ("<< numerator2 <<"/"<< denominator2 <<") = " << c << "/" << d;
    cout << "\n";
    return 0;
}
 
 
 
int main()         
{
    multiply();
    system ("PAUSE");
}

---

Why c and d its 0??????

Why are you not listening?
Go read up on the command operator, passing by value and passing by reference.

its working:):) thanks