使用循环语句在 C 中练习:编写代码以使用 do-while 语句反转数字

Exercise in C with loop statements: Write code to reverse numbers with do-while statement

我有一个任务(目前正在研究循环语句,所以我处于初学者阶段)要求编写一个程序来反转整数,因此它必须具有 do 语句。

输出应该是(例子):

Enter a number: 4568
The reversal is: 8654

请记住,由于到目前为止我一直在关注我的书,所以我已经学习并了解了非常基础的知识 + 选择和循环语句。我的选择非常有限,所以 no 数组。

这本书建议制作一个 do 循环,将数字重复除以 10,直到达到 0 这是我目前所做的(代码未完成):

int main(void)
{
  int n,x;

  printf("Enter a number: ");
  scanf("%d", &n);

  printf("The reversal is: ");

  x = n % 10;

  printf("%d",x); /*outputs the last number digit in the first place*/ 

  do{



   ....
   n  /= 10;   /* for example if I divide the number 56222  by ten the output would come out as 
               5622 562 56 5*/
   ....

  }while (n!=0);



  return 0;
}

如您所见,我找到了一种将最后一位数字放在首位的方法,但是 在用这个非常有限的选择将数字除以 10 之后,我正在努力弄清楚如何反转其余数字。

我应该在 do 语句中输入什么? 非常感谢。


int main(void)
{
      
    int n,x;
    
    printf("Enter a number: ");
    scanf("%d", &n);
    
    printf("The reversal is: ");
    
    int rev = 0;     
    
    do{
      
        x = n % 10;
        rev = rev * 10 + x; 
        n  /= 10;
    
    }while (n!=0);
    
    printf ("%d", rev);
 
    return 0;
    
}

here you need a new integer rev whose value is 0 initially. Lets take example of 432

n = 432

when you do x = n % 10 x = 2

so when you do rev = rev * 10 + x rev is 0 and value of rev will be 2

n /= 10 make n = 43

-->so in the next iteration

n = 43 x = n % 10 results in x = 3

rev value now is 2

so rev = rev * 10 + x results in 2 * 10 + 3

so rev = 23

n /= 10 results in n = 4

-->in the last iteration n = 4 x = n % 10 results in x = 4 rev value now is 23

so rev = rev * 10 + x results in 23 * 10 + 4 so rev = 234

n /= 10 results in n = 0

so when you print rev you get the answer 234 reverse of 432