如何调用简化 C 中分数的函数？

Question

这是我需要做的：

定义一个名为“fraction”的结构，其中包含整数成员“numerator”和 “分母。”提示用户以“#/#”的形式输入分数。调用一个名为 “simplFrac” 通过将分子和分母除以最大的分数来简化分数公因数和 returns 调用函数的简化分数。打印结果来自主要（）。

这是我的代码：

  #include <stdio.h>

  struct fraction {
    int numerator;
    int denominator;
  };

  struct fraction simplFrac(struct fraction x);

 int main(void) {

  struct fraction a;

  printf("Enter a fraction in the form #/#: ");
  scanf("%i/%i",&a.numerator,&a.denominator);

  printf("\n");

  printf("The simplified fraction is: %i/%i\n",);

  return 0;
 }

struct fraction simplFrac(struct fraction x) {
  
  int gcf,remainder;

  while (x.numerator != 0)
   {
      remainder = x.denominator % x.numerator;
      x.denominator = x.numerator;
      x.numerator = remainder;
   }

   gcf = x.denominator;

   x.numerator = x.numerator / gcf;
   x.denominator = x.denominator / gcf;

   return x;
}

Answer 1

请您尝试以下操作：

#include <stdio.h>

struct fraction
{
    int numerator;
    int denominator;
};

struct fraction simplFrac(struct fraction x);
int gcf(int, int);

int main(void)
{
    struct fraction a;
    char buf[BUFSIZ];

    printf("Enter a fraction in the form #/#: ");
    fgets(buf, BUFSIZ, stdin);
    sscanf(buf, "%i/%i", &a.numerator, &a.denominator);

    a = simplFrac(a);

    printf("The simplified fraction is: %i/%i\n", a.numerator, a.denominator);

    return 0;
}

struct fraction simplFrac(struct fraction x)
{
    int factor = gcf(x.denominator, x.numerator);

    x.numerator /=  factor;
    x.denominator /= factor;

    return x;
}

int gcf(int a, int b)
{
    if (b == 0) return a;
    else return gcf(b, a % b);
}

计算gcf的算法被分离为一个函数gcf()。

如何调用简化 C 中分数的函数？

How to call a function that simplifies a fraction in C?

c

struct

function

fractions

data-structures