了解一些使用循环扭转 [英] Understanding number reversing using loop

问题描述

我钻进在C编程这个简单的程序，该程序是简单地扭转由用户任意输入数字，使用whil​​e循环，我不知道它的好张贴这样的问题，但我没有真的知道如何while循环工程

  INT N，反向= 0;   的printf（请输入一个数字，以扭转\\ n）;
   scanf函数（％d个，＆安培; N）;   而（N！= 0）
   {
      反向=反向* 10;
      反向=反转+ N％10;
      N = N / 10;
   }   的printf（输入数字的反向是=％d个\\ N，反向）;
 

我会如此心存感激，如果任何人都可以解释给我

解决方案

 而（条件为真）
{
    // 做东西。
    //通常情况下，但并非总是如此，在东西我们正在做的是一件修改
    //我们正在检查的条件。
}
 

所以你的情况只要 N 不等于 0 ，你会继续在内容执行循环。如果 0 在 scanf函数输入（）循环会被一起跳过。

在你每次循环你使用整数除法的属性，以使值 N 由10即较小的因素：

  N = 541
N = N / 10 =一十分之五百四十一= 54
N = N / 10 = 54/10 = 5
N = N / 10 = 5/10 = 0
 

所以 N 最终将是0，循环将退出。

i got into this simple program in c programing, the program is simply reversing any input number by user, using a while loop, i dont know if its okay to post such a question but i didnt realy understood how the while loop works

int n, reverse = 0;

   printf("Enter a number to reverse\n");
   scanf("%d",&n);

   while (n != 0)
   {
      reverse = reverse * 10;
      reverse = reverse + n%10;
      n = n/10;
   }

   printf("Reverse of entered number is = %d\n", reverse);

i would be so thankfull if anyone could explain it to me

解决方案

while(condition is true)
{
    // Do stuff.
    // Normally, but not always, the "stuff" we're doing is something that modifies
    // the condition we're checking.
}

So in your case as long as the content of n is not equal to 0, you will continue to execute the loop. If 0 is entered in the scanf() the loop will be skipped altogether.

At each iteration of your loop you're using the property of integer division to make the value of n smaller by a factor of 10. ie:

n = 541
n = n / 10 = 541/10 = 54
n = n / 10 = 54/10  = 5
n = n / 10 = 5/10   = 0 

So n will eventually be 0 and the loop will exit.

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