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("\nFractional part1: %s \n",fractionToDecimal(n, d));
   printf("enter numerator2 and denominator2:");
   scanf("%d%d",&n,&d);
   printf("\nn = %d, d = %d ", n, d);
   printf("\nFractional part2: %s\n ",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)

Bhanu Priya

Updated on 03-Sep-2021 06:42:09

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