4

假设这些函数已经给出：

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;
}

我的任务是写GCD()and add_fractions()，这就是我写的：

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;
}

假设目前两个分母相等，为什么我不能用 Cygwin 运行它？我用这个命令编译

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

我有两个错误：（Cygwin 在我的电脑上是西班牙语，所以我不确定我要写的是准确的翻译）：

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

（分母也一样）

问题是什么？

4

3 回答 3

5 
const struct FRACTION *a, const struct FRACTION *b

所以a和b是指向常量的指针struct FRACTION。然后你写：

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

您不使用 访问结构指针的成员.，但使用->，这应该是

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

附言 1：括号不是必须的，不要多加，会降低可读性。

附言 2：你的加法公式坏了，用

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

反而。

于 2012-12-06T16:35:02.813 回答
1

在您的函数中add_fractions，您已将a和声明b为指向struct FRACTION：

struct FRACTION add_fractions(const struct FRACTION *a, const struct FRACTION *b)

因为aandb是指针，而不是结构，所以你不能写像a.numerator.

您需要取消引用 a并且b在使用.运算符之前：

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

或使用->组件选择运算符，这是做同样事情的简写方式：

c.numerator = (a->numerator) + (b->numerator);
c.denominator = a->denominator ;
于 2012-12-06T16:37:32.307 回答
1

除了user529758你的加法算法应该是

  n1 * d2 + n2 * d1
= ------------------
       d1 * d2

我认为需要强调乔纳森的观点，然后您需要将结果的分子和分母除以 ，GCD以将结果分数减少到最低形式。

例如，您可以实现一个新函数来进行归约：

struct FRACTION reduce_fraction(const struct FRACTION *frac)
{
    struct FRACTION reduced;
    int gcd = GCD(frac->numerator, frac->denominator);

    reduced.numerator = frac->numerator / gcd;
    reduced.denominator = frac->denominator / gcd;

    return reduced;
}

您的add_fraction函数现在将如下所示：

struct FRACTION add_fractions(const struct FRACTION *a, const struct FRACTION *b)
{                                    
    struct FRACTION c;

    c.numerator = (a->numerator * b->denominator) + (b->numerator * a->denominator);
    c.denominator = a->denominator * b->denominator;
    return reduce_fraction(&c);
}

此外，鉴于您已经完成了艰苦的工作来确定评论中的结果，而不是printfing to stdout，为什么不assert这样做 - 这为自动化单元测试提供了基础：

#include <assert.h>
...
m = 15; n = 18;
assert(GCD(m, n) == 3);
assert(GCD(m, n) == GCD(n,m));

m = 80; n = 20;
assert(GCD(m, n) == 20);
assert(GCD(m, n) == GCD(n,m));

和 也add_fractions一样reduce_fraction:)

于 2014-07-27T08:05:11.393 回答