# C program to represent numbers in numerator and denominator in string format

## Problem

Find the fraction part from two given integers given by user at run time by using the dynamic memory allocation and represent the numerator and denominator in string format.

## Solution

The solution to represent the numerator and denominator in string format is as follows −

Example -

• The input is given below −
Numerator1 = 3
Denominator2 = 2

numerator2 = 4
denominator2 = 7
• The output is as follows −
Fractional part1: 1.5
Fractional part2: 0.(571428)

## Example

Following is the C program to represent the numerator and denominator in string format

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
char* fractionToDecimal(int numerator, int denominator) {
char *p;
int psz, n, *dec, dsz, x;
long long num, den, k, f;
int i, repeat_at;
int neg = 0;
psz = dsz = 100; n = x = 0;
p = malloc(psz * sizeof(char));
//assert(p);
neg = ((numerator > 0 && denominator < 0) ||
(numerator < 0 && denominator > 0)) ? 1 : 0;
num = numerator;
den = denominator;
num = (num < 0) ? -num : num;
den = (den < 0) ? -den : den;
k = num / den;
f = num % den;
if (neg && (k || f)) p[n ++] = '-';
n += sprintf(&p[n], "%lld", k);
if (!f) {
p[n] = 0;
return p;
}
p[n ++] = '.';
dec = malloc(dsz * sizeof(int));
repeat_at = -1;
if (f < 0) f = -f;
while (f) {
for (i = 0; i < x; i += 2) {
if (dec[i] == f) {
repeat_at = i;
goto done;
}
}
if (x + 1 >= dsz) {
dsz *= 2;
dec = realloc(dec, dsz * sizeof(int));
}
dec[x ++] = f;
f *= 10;
k = f / den;
dec[x ++] = k;
f = f % den;
}
done:
for (i = 0; i < x; i += 2) {
if (n + 3 > psz) {
psz *= 2;
p = realloc(p, psz * sizeof(char));
}
if (repeat_at == i) {
p[n ++] = '(';
}
p[n ++] = '0' + dec[i + 1];
}
if (repeat_at != -1) p[n ++] = ')';
p[n ++] = 0;
free(dec);
return p;
}
int main(void){
int n,d;
printf("enter numerator1 and denominator1:");
scanf("%d%d",&n,&d);
printf("n = %d, d = %d ", n, d);
printf("Fractional part1: %s ",fractionToDecimal(n, d));
printf("enter numerator2 and denominator2:");
scanf("%d%d",&n,&d);
printf("n = %d, d = %d ", n, d);
printf("Fractional part2: %s ",fractionToDecimal(n, d));
return 0;
}

## Output

When the above program is executed, it produces the following result −

enter numerator1 and denominator1:4 5
n = 4, d = 5
Fractional part1: 0.8
enter numerator2 and denominator2:5 9

n = 5, d = 9
Fractional part2: 0.(5)