C - 具有多个案例编号的开关

Question

所以我的教授要求我们创建一个switch语句.我们只允许使用"SWITCH"语句来执行该程序.他希望我们输入一个数字然后显示它,如果它在数字范围内,将显示如下所示的公文包编号.现在......我知道对于这种类型的程序,使用IF语句更容易.做案例1:案例2:案例3 ......案例30将起作用,但由于数字范围将花费太多时间.

#include <stdio.h>
main()
{
      int x;
      char ch1;
      printf("Enter a number: ");
      scanf("%d",&x);
      switch(x)
      {
                 case 1://for the first case #1-30
                 case 30:
                      printf("The number you entered is >= 1 and <= 30");
                      printf("\nTake Briefcase Number 1");
                      break;         
                 case 31://for the second case #31-59
                 case 59:
                      printf("The number you entered is >= 31 and <= 59");
                      printf("\nTake Briefcase Number 2");
                      break;                 
                 case 60://for the third case #60-89
                 case 89:
                      printf("The number you entered is >= 60 and <= 89");
                      printf("\nTake Briefcase Number 3");
                      break;                 
                 case 90://for the fourth case #90-100
                 case 100:
                      printf("The number you entered is >= 90 and <= 100");
                      printf("\nTake Briefcase Number 4");
                      break;      
                 default:
                     printf("Not in the number range");
                     break;

                 }
      getch();
      }

我的教授告诉我们,如何做到这一点有一个较短的方法,但不会告诉我们如何做.我能想到缩短它的唯一方法是使用IF,但我们不允许这样做.关于如何使这项工作成功的任何想法？

Answer 1

使用GCC和CLang,您可以使用案例范围,如下所示:

switch (x){

case 1 ... 30:
    printf ("The number you entered is >= 1 and <= 30\n");
    break;
}

唯一的交叉编译器解决方案是使用这样的case语句:

switch (x){

case 1:
case 2:
case 3:
case 4:
case 5:
case 6:
    printf ("The number you entered is >= 1 and <= 6\n");
    break;
}

编辑:使用某些东西来实现switch (x / 10)这一点是另一种好方法.当范围不是差异时,使用GCC案例范围可能更简单10,但另一方面,您的教授可能不会将GCC扩展作为答案.

Answer 2

如果范围一致,那么您可以丢弃一些数据:

switch (x / 10 )
{
   case 0:
   case 1:
   case 2:  // x is 0 - 29
     break ;

   // etc ...
}

否则你将不得不在边缘做一些hackery.

Answer 3

   Try this ...

#include <stdio.h>
main() {
      int x;
      char ch1;
      printf("Enter a number: ");
      scanf("%d",&x);
      int y=ceil(x/30.0);
      switch(y) {
                 
                 case 1:
                      printf("The number you entered is >= 1 and <= 30");
                      printf("\nTake Briefcase Number 1");
                      break;         
                 
                 case 2:
                      printf("The number you entered is >= 31 and <= 60");
                      printf("\nTake Briefcase Number 2");
                      break;                 
                
                 case 3:
                      printf("The number you entered is >= 61 and <= 90");
                      printf("\nTake Briefcase Number 3");
                      break;                 
                 
                 case 4:
                      printf("The number you entered is >= 91 and <= 100");
                      printf("\nTake Briefcase Number 4");
                      break;      
                 default:
                     printf("Not in the number range");
                     break;

                 }
      getch();
      }