Hi all,

I am working on a C++ program that converts a decimal to a fraction, finds the lowest common multiple, multiplies the numerator by the denominator, then reduces the fraction again so the fraction will be 'numerator/1'.

I am using this code in a larger program to change a floating point slope value to a whole value for slope-intercept equations and standard form equations.

The program will run but I tested almost 500 numbers and 20 numbers made the program crash (it stalled and did not finish the equation, I had to close the terminal and open it up again... )

These 20 numbers I had problems with:

2.36, 7.36, 1.11, 1.001, 1.66, 1.88, 2.22, 2.13, 3.24, 3,26, 3.32, 3.43, 3.49, 3.51, 3.57, 3.68, 3.74, 3.76, 3.82

(I am sure there are more...)

I have tried c-style casting and dynamic and static casting to try to resolve this issue. Can someone help me out and point out why these certain numbers make the program crash and offer a possible sloution?

I tried debugging this program on gdb and keep on getting this frustrating message..

(gdb) break 16
No line 16 in file "init.c".
(gdb)


Here is the program's code:

//example.cc
#include <iostream>
#include "reduce.h"
#include "decimal-to-fraction.h"
using namespace std;

int main()
{
    double deci;
    double n, d;

    while(1) {
        cout << "Enter a decimal to convert to a fraction ('0' to quit): ";
        cin >> deci;
        cin.ignore(INT_MAX, '\n');
        if(!deci) exit(0);
        
        dec_to_frac(deci, n, d);
        
        if (n * d != n){
            cout << '\n' << deci << " = " << n << '/' << d << "\n\n";
            cout<<'\n' << d << " x " << n;
            n = d * n;
            cout<< " = " << n << "\n\n";
            cout<<'\n' << n << '/' << d; 
            reduce(n, d);
        
            cout << " = " << n << '/' << d << "\n\n";

            cout<<'\n' << n << "\n\n";
        }
        
        else
            cout<<'\n' << deci<< "\n\n";
    
    }

    return 0;
}

#ifndef _REDUCE_H_

#error You must include "reduce.h" before this file

#endif /* def _REDUCE_H_ */



#ifndef _DECIMAL_TO_FRACTION_H_

#define _DECIMAL_TO_FRACTION_H_



void dec_to_frac(double decimal, double &numerator, double &denominator)

{

    //numerator = static_cast<int >(decimal);
    //numerator = (int )decimal;
    numerator = decimal;

    denominator = 1;
    


    while(numerator != (int)numerator) {

        numerator *= 10;

        denominator *= 10;

    }

    reduce(numerator, denominator);

}



#endif /* def _DECIMAL_TO_FRACTION_H_ */

#ifndef _REDUCE_H_

#define _REDUCE_H_



void reduce(double &numer, double &denom)

{

    int i;



    for(i=2; i<=numer; ++i) {
        

        if( ((numer/i) == ((int )(numer/i))) && ((denom/i) == ((int)(denom/i))) ) {

            numer /= i;

            denom /= i;

            --i;              

        }

    }

}



#endif /* def _REDUCE_H_ */

You should post ALL your custom includes as another "code" entry in text, just like the main program.
Nobody likes to open attachments...

Did you try removing the breakpoint in question?
sometimes it's just as easy or easier to insert a "test cout" that reports the state of variables in a non-cooperative program.

Comments
Hell yeah, attachments suck-ass

There are many ways to do this:

One is to use the fact that any fractional part can be represented by:-

x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + \frac{1}{\ddots\,}}}}

where a0 is some integer and all the other numbers an are positive integers. (continued fractions)

This elegant idea is described quite nicely here

The other way is to look at it is as follows. Count the number of digits after the decimal place. 0.345 In this case it is three digits after the decimal place. Let the number of zeros after the decimal place be the number you divide by with a one in front of it i.e = 1000 (three zeros) so you can apply Euclids GCD algo on 345/1000 .

1.93 is two (100). So GCD(93/100) etc.

0.04 is two. So GCD(4/100) etc.

This article has been dead for over six months. Start a new discussion instead.