c - 缩短长代码

Question

人!

我被告知要编写下一个代码作为家庭作业。 如果你编译它 - 你很容易看到它的目的。现在，我的问题是是否有办法让它更短(我是 C 的新手)。我必须使用结构和结构指针。这似乎是一个蹩脚的问题 - 对此感到抱歉。 另外，我想知道是否可以重复调用“main()”。

#include <stdio.h>

typedef struct frac{
    int num;
    int den;
};

int reducer( struct frac *fi ){
    if( fi->num == 0 ) return 0;
    if( fi->den == 1 ) return 1;
    if( fi->num % fi->den == 0 ){
        fi->num /= fi->den;
        fi->den /= fi->den;
        return reducer( fi );
    }
    if( fi->num % 2 == 0 && fi->den % 2 == 0 ){
        fi->num /= 2;
        fi->den /= 2;
        return reducer( fi );
    }
    else if( fi->num % 3 == 0 && fi->den % 3 == 0 ){
        fi->num /= 3;
        fi->den /= 3;
        return reducer( fi );
    }
}

int main(){
    char c , tt;
    struct frac one , two , multi , quot , sum , diff , *o , *t , *m , *q , *s , *d;
    printf( "Please, enter the first fraction, ieg. 3/8:\n" );
    scanf( "%d/%d%c" , &one.num , &one.den , &tt );
    printf( "Now the second fraction (numerator/denominator):\n" );
    scanf( "%d/%d%c" , &two.num , &two.den , &tt );
    o = &one;
    t = &two;
    m = &multi;
    q = &quot;
    s = &sum;
    d = &diff;
    m->num = o->num * t->num; // product numerator
    m->den = o->den * t->den; // product denominator
    q->num = o->num * t->den; // quotient numerator
    q->den = o->den * t->num; // quotient denominator and so on...
    s->num = q->num + q->den;
    s->den = m->den;
    d->num = q->num - q->den;
    d->den = m->den;
    reducer( q );
    reducer( m );
    reducer( s );
    reducer( d );
    printf( "%d/%d + %d/%d = %d/%d\n" , o->num , o->den , t->num , t->den , s->num , s->den );
    printf( "%d/%d - %d/%d = %d/%d\n" , o->num , o->den , t->num , t->den , d->num , d->den );
    printf( "%d/%d * %d/%d = %d/%d\n" , o->num , o->den , t->num , t->den , m->num , m->den );
    printf( "%d/%d : %d/%d = %d/%d\n" , o->num , o->den , t->num , t->den , q->num , q->den );
    printf( "\nWould you like to make another calculation? (y/n):\n" );
    scanf( "%c" , &c );
    if( c == 121 || c == 89 ){
        return main();
    }
    return 0;
}

Answer 1

这里有一些建议:

• 使用循环而不是递归。在这种情况下它更自然，并且不会使堆栈随着每次迭代而增长:

  int finish;
  do
  {
      //...
      printf( "\nWould you like to make another calculation? (y/n):\n" );
      scanf( "%c" , &c );
      finish = c != 121 && c == 89;
  }
  while (!finish)
  

• 您可以删除所有指针声明并直接使用分数本身。

  multi.num = one.num * two.num; // product numerator
  // ...
  reducer(&quot)
  

希望对您有所帮助!