Question

我正在学习决赛，并担心我在这个话题上花了太多时间，所以我想要一些意见。这将是一个长篇大论，所以我希望有人看到这一点。以下问题出现在“从初学者到专业人士”的书中：

“定义一个名称为Length的结构类型，表示以码为单位的长度，英尺，和英寸。定义一个add（）函数，它将添加两个Length参数并返回总和为长度类型。定义第二个函数show（），它将显示该值其长度参数。编写一个将使用Length类型和add（）的程序和show（）函数以码，英尺和英寸的形式对任意数量的长度求和从键盘输入并输出总长度。“

我的解释如下：

#include "stdio.h"
#include "stdlib.h"
#include"string.h"

struct length
{
    float FT;
    float YD;
    float IN;

};

struct length ReadIT(char sym[])
{
    struct length Convert;
    float dummy;
    printf("Enter %s\n",sym);
    scanf("%f",&dummy);
    Convert.FT = dummy;
    Convert.YD = dummy/3;
    Convert.IN = dummy*12;
    return(Convert);
};
struct length Add(struct length first, struct length second)
{
    struct length total;
    total.FT = first.FT+second.FT;
    total.YD = first.YD + second.YD;
    total.IN = first.IN + second.IN;
    return total;
};
void Show(struct length Convert )
{
    printf("Conversions FT: %.2f\n",Convert.FT );
    printf("Conversions YD: %.2f\n",Convert.YD );
    printf("Conversions IN: %.2f\n",Convert.IN );
}
void ShowSum(struct length total)
{
    printf("total FT: %.2f\n",total.FT );
    printf("total YD: %.2f\n",total.YD );
    printf("total IN: %.2f\n",total.IN );
}


int main(void)
{

char cmd = 'n';
struct length L1,L2;
do
{

    L1 = ReadIT("Length 1");
    Show(L1);
    L2 = ReadIT("Length 2");
    Show(L2);
    Add(L1,L2);
    ShowSum(Add(L1,L2));//adds up the two lengths , but how do I store them?
    printf("Would you like to add more lengths? Type 'y' to continue");            

    scanf("%s",&cmd);
    }
    while(tolower(cmd)=='y');
    printf("The total of all lengths is : ");

    return 0;
}

另外，因为这本书出来了：我发现我应该像这样解释它：

#include <stdio.h>
#include <ctype.h>

#define INCHES_PER_FOOT 12
#define FEET_PER_YARD    3

struct Length
{
  unsigned int yards;
  unsigned int feet;
  unsigned int inches;
};

struct Length add(struct Length first, struct Length second);
void show(struct Length length);

int main(void)
{
  char answer = 'n';
  struct Length length;
  struct Length total = { 0,0,0};
  int i = 0;
  do
  {
    printf("Enter a length in yards, feet, and inches: ");
    scanf(" %d %d %d", &length.yards, &length.feet, &length.inches);
    total = add(total,length);
    printf("Do you want to enter another(y or n)?: ");
    scanf(" %c", &answer);
    fflush(stdin);
   }while(tolower(answer) == 'y');
  printf("The total of all the lengths is: ");
  show(total);
  printf("\n");
  return 0;
}

struct Length add(struct Length first, struct Length second)
{
  unsigned long inches = 0;
  struct Length sum;
  inches = first.inches + second.inches+
    INCHES_PER_FOOT*(first.feet+second.feet+FEET_PER_YARD* (first.yards+second.yards));
  sum.inches = inches%INCHES_PER_FOOT;
  sum.feet = inches/INCHES_PER_FOOT;
  sum.yards = sum.feet/FEET_PER_YARD;
  sum.feet %= FEET_PER_YARD;
  return sum;
}

void show(struct Length length)
{
  printf("%d yards %d feet %d inches", length.yards,length.feet,length.inches);     
}

当我看到这个答案时，我的第一个想法是，“我本可以做到这一点”，但这听起来好像他们希望所有这两个长度分别在每两个输入中完成，以及单个FT，YD的总数，＆amp; IN独立完成。现在我痴迷于完成我的版本，但我已经碰壁了。有人可以让我去吗？

Answer 1

我击中的墙是我无法想到一种方法来存储总数首先循环，然后将其添加到下一循环的总数中把它作为一个变量存储，然后重复这个只要用户想要，就会一遍又一遍。

你可以做的非常类似于本书的解决方案：

  struct Length total = { 0, 0, 0 }, subtotal;
…
    ShowSum(subtotal = Add(L1, L2)); // adds up the two lengths and stores the sum
    total = add(total, subtotal);

偏离指示，有趣但信息丰富的结构练习

1 个答案: