如何反转以零C＃结尾的数字 [英] How to reverse a number ending with zero C#

问题描述

i低于代码

Hi,

i have below code

fiveDigNo=17660;

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

我希望像06671一样反转。但我得到6671.



我的尝试：



反转功能。

I want to get reverse like 06671. But i am getting 6671.

What I have tried:

Reverse Function .

推荐答案

这是一样的。 06671 = 6671.这不是很明显吗？



如果你想获得06671，那不是关于你的号码，这是关于你呈现的方式数字作为字符串：

Int32.ToString方法（系统） [ ^ ]，

Int32.ToString方法（字符串）（系统） [ ^ ]，

标准数字格式字符串 [ ^ ]，

自定义数字格式字符串 [ ^ ]。



严格来说，问题是反转一个数字没有正确定义。例如，如果您反转数字的八进制，十六进制数字等，您将得到不同的结果。所以，或者，您可以先通过 ToString 或以其他方式获取字符串，然后反转该字符串。

-SA

This is the same. 06671 = 6671. Isn't it obvious?

If you want to obtain "06671", it it not about your number, this is about the way you present the number as a string:
Int32.ToString Method (System)[^],
Int32.ToString Method (String) (System)[^],
Standard Numeric Format Strings[^],
Custom Numeric Format Strings[^].

Strictly speaking, the problem "reverse a number" wasn't correctly defined. For example, you would get different result if you "reverse" the number's octal, hexadecimal digits, and so on. So, you could, alternatively, simply get a string first, via ToString or in some other way, and then reverse that string.

—SA

'06671'和' 6671'是相同整数的两个不同的字符串表示。也就是说，如果您确实需要输出中的前导 0 ，那么您必须迟早将整数变量更改为字符串1。

'06671' and '6671' are two different string representations of the same integer number. That is, if you really need the leading 0 in the output then you have to change your integer variable into a string one, sooner or later.

数字06671和6671的数字相同。如果你想要一个像06671（字符串表示）的结果，你必须使用一个字符串并反转：

The numbers 06671 and 6671 are numerical identical. If you want a result like "06671" (a string representation) you have to use a string and reverse that:

string numString = fiveDigNo.ToString();
char[] charArray = numString.ToCharArray();
Array.Reverse(charArray);
string revString = new string(charArray);
// If you need the numeric value too:
int revNumber = Convert.ToInt32(revString);

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