c-偏离指示，有趣但信息丰富的结构练习

Question

我正在为期末考试做准备，担心我在这个话题上花了太多时间，所以我想要一些意见。这将是一个很长的阅读，所以我希望有人看到这一点。以下问题出现在书中，"从新手到专业人士的C阶段"：

"定义一个结构类型，其名称为Length，表示以码、英尺、，和英寸。定义一个add（）函数，该函数将添加两个Length参数并返回总和为类型Length。定义第二个函数show（），它将显示值的Length参数。编写一个使用Length类型和add（）的程序和show（）函数用于对任意数量的长度（以码、英尺和英寸为单位）求和从键盘输入并输出总长度。"

我这样解释：

#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 %sn",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: %.2fn",Convert.FT );
    printf("Conversions YD: %.2fn",Convert.YD );
    printf("Conversions IN: %.2fn",Convert.IN );
}
void ShowSum(struct length total)
{
    printf("total FT: %.2fn",total.FT );
    printf("total YD: %.2fn",total.YD );
    printf("total IN: %.2fn",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;
}

此外，自从这本书问世以来：我发现我应该这样解释它：

#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，"我本可以这么做的"，但这听起来像是他们希望每两个输入都单独完成长度的相加，以及单个FT、YD和&IN独立完成。现在我痴迷于完成我的版本，但我遇到了困难。有人能让我走吗？

Answer 1

我碰到的墙是，我想不出一种方法来存储首先循环，然后将其添加到下一个循环的总数中大约，将总数存储为一个变量，然后重复此操作一次又一次，只要用户想要。

你可以像书中的解决方案一样完成它：

  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);