How to add two fractions?

Question

Assuming that these functions are already given:

include <stdio.h>     /* printf                         */
include "fractions.h" /* struct FRACTION, add_fractions */

struct FRACTION make_fraction(int numerator, int denominator)
{
  struct FRACTION f;

  f.numerator = numerator;
  f.denominator = denominator;

  return f;
}

void test_fraction(int numerator1, int denominator1, 
                   int numerator2, int denominator2)
{
  struct FRACTION a = make_fraction(numerator1, denominator1);
  struct FRACTION b = make_fraction(numerator2, denominator2);

  struct FRACTION c = add_fractions(&a, &b);

  printf("%i/%i + %i/%i = %i/%i\n", a.numerator, a.denominator,
                                    b.numerator, b.denominator,
                                    c.numerator, c.denominator);
}

void testGCD(void)
{
  int m, n;

  m = 15; n = 18; /* GCD is 3 */
  printf("GCD of %i and %i is %i\n", m, n, GCD(m, n));
  printf("GCD of %i and %i is %i\n", n, m, GCD(n, m));

  m = 80; n = 20; /* GCD is 20 */
  printf("GCD of %i and %i is %i\n", m, n, GCD(m, n));
  printf("GCD of %i and %i is %i\n", n, m, GCD(n, m));

  m = 21; n = 47; /* GCD is 1 */
  printf("GCD of %i and %i is %i\n", m, n, GCD(m, n));
  printf("GCD of %i and %i is %i\n", n, m, GCD(n, m));

  m = 68; n = 153; /* GCD is 17 */
  printf("GCD of %i and %i is %i\n", m, n, GCD(m, n));
  printf("GCD of %i and %i is %i\n", n, m, GCD(n, m));
}

int main(void)
{
  testGCD();

  test_fraction(2, 3, 1, 6);
  test_fraction(1, 5, 4, 9);
  test_fraction(3, 7, 12, 21);
  test_fraction(5, 8, 3, 16);
  test_fraction(7, 8, 3, 12);
  test_fraction(0, 8, 3, 16);
  test_fraction(1, 1, 3, 16);
  test_fraction(5, 8, -3, 16);
  test_fraction(1, 5, -4, 9);
  test_fraction(-1, 5, -4, 9);


  return 0;
}

My task is to write GCD() and add_fractions(), and this is what I have written:

include "fractions.h" 

struct FRACTION add_fractions(const struct FRACTION *a, const struct FRACTION *b)
{                                    
    struct FRACTION c ; /*result struct*/
   /* int GCD_a = GCD(a.numerator, a.denominator);  GCD of the fraction a*/ 
    /*int GCD_b = GCD(b.numerator, b.denominator);  GCD of the fraction b*/

    c.numerator = (a.numerator) + (b.numerator);
    c.denominator = a.denominator ;


    return c;
  /* struct FRACTION empty;*/
   /*return empty;*/
}

int GCD(int a, int b)
{
    /*Variables*/
    int remainder = 0; /*remainder*/
    int larger = a;    
    int smaller = b;

    remainder = larger % smaller;
    while (remainder != 0)
    {
      larger = smaller;
      smaller = remainder;
      remainder = larger % smaller;
    }

    return smaller;
}

Assuming that for the moment both denominators are equals, why I can't run this with Cygwin? I use this command to compile

gcc -Wall -Wextra -ansi -pedantic -Wno-unused-parameters main.c fractions.c -o fractions.exe    

and I have two errors: (Cygwin is in Spanish on my computer so I am not sure that what I am going to write is the exact translation):

error: trying to put "numerator" in something which is not a struct

(and the same for denominator)

What is the problem?

Answer 1

const struct FRACTION *a, const struct FRACTION *b

So a and b are pointers to a constant struct FRACTION. Then later you write:

c.numerator = (a.numerator) + (b.numerator);

you don't access members of struct pointers using ., but using ->, this should be

c.numerator = a->numerator + b->numerator;

P. s. 1: the parentheses are not needed, don't put them in superfluously, they decrease readability.

P. s. 2: your addition formula is broken, use

c.numerator = a->numerator * b->denominator + b->numerator * a->denominator;
c.denominator = a->denominator * b->denominator;

instead.