blocks|key|2277298|text|using+System;+

public+class+DoWhileDemo+{+++
++public+static+void+Main()+{+
++++int+num;+
++++int+nextdigit;+

++++num+=+198;+

++++Console.WriteLine("Number:+"+%2B+num);+

++++Console.Write("Number+in+reverse+order:+");+

++++do+{+
++++++nextdigit+=+num+%25+10;+
++++++Console.Write(nextdigit);+
++++++num+=+num+/+10;+
++++}+while(num+>+0);+

++++Console.WriteLine();+
++}+++
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2277299|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>using System; 

public class DoWhileDemo { 
 public static void Main() { 
 int num; 
 int nextdigit; 

 num = 198; 

 Console.WriteLine("Number: " + num); 

 Console.Write("Number in reverse order: "); 

 do { 
 nextdigit = num % 10; 
 Console.Write(nextdigit); 
 num = num / 10; 
 } while(num &gt; 0); 

 Console.WriteLine(); 
 } 
}
</code></pre>

blocks|key|2000952|text|像这样的东西？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2000953|public+int+ReverseInt(int+num)
{
++++int+result=0;
++++while+(num>0)+
++++{
+++++++result+=+result*10+%2B+num%2510;
+++++++num+/=+10;
++++}
++++return+result;
}|code-block|syntax|javascript|2000954|作为一个粗俗的一行(更新:使用本杰明的评论来缩短它)：|2000955|num.ToString().Reverse().Aggregate(0,+(b,+x)+=>+10+*+b+%2B+x+-+'0');|2000956|一个速度更快的1.25行代码：|2000957|public+static+int+ReverseOneLiner(int+num)
{
++++for+(int+result=0;;+result+=+result+*+10+%2B+num+%25+10,+num+/=+10)+if(num==0)+return+result;
++++return+42;
}|2000958|这不是一行，因为我必须包含return+42;。C#编译器不让我编译，因为它认为没有代码路径返回值。|offset|length|style|CODE|2000959|附注:如果你写了这样的代码，而一个同事抓住了它，那么他/她对你所做的一切都是你应得的。请注意！|2000960|编辑:我想知道LINQ一行程序的速度有多慢，所以我使用了以下基准代码：|2000961|public+static+void+Bench(Func<int,int>+myFunc,+int+repeat)
{
++++var+R+=+new+System.Random();
++++var+sw+=+System.Diagnostics.Stopwatch.StartNew();
++++for+(int+i+=+0;+i+<+repeat;+i%2B%2B)
++++{
++++++++var+ignore+=+myFunc(R.Next());
++++}
++++sw.Stop();
++++Console.WriteLine("Operation+took+{0}ms",+sw.ElapsedMilliseconds);
}|2000962|结果(10%5E6个随机数在正int32范围内)：|2000963|While+loop+version:
Operation+took+279ms

Linq+aggregate:
Operation+took+984ms|2000964|entityMap^0|0|0|0|0|0|0|D|A|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1C|8|@$Q|1D|R|1E|S|T]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1G|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|10|3|11|5|6|7|1I|8|@]|9|@]|A|$]]|$1|12|3|13|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|14|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|15|$]]

Something like this?

<pre><code>public int ReverseInt(int num)
{
 int result=0;
 while (num&gt;0) 
 {
 result = result*10 + num%10;
 num /= 10;
 }
 return result;
}
</code></pre>

As a hackish one-liner (update: used Benjamin's comment to shorten it):

<pre><code>num.ToString().Reverse().Aggregate(0, (b, x) =&gt; 10 * b + x - '0');
</code></pre>

A speedier one-and-a-quarter-liner:

<pre><code>public static int ReverseOneLiner(int num)
{
 for (int result=0;; result = result * 10 + num % 10, num /= 10) if(num==0) return result;
 return 42;
}
</code></pre>

It's not a one-liner because I had to include <code>return 42;</code>. The C# compiler wouldn't let me compile because it thought that no code path returned a value.

P.S. If you write code like this and a co-worker catches it, you deserve everything he/she does to you. Be warned!

EDIT: I wondered about how much slower the LINQ one-liner is, so I used the following benchmark code:

<pre><code>public static void Bench(Func&lt;int,int&gt; myFunc, int repeat)
{
 var R = new System.Random();
 var sw = System.Diagnostics.Stopwatch.StartNew();
 for (int i = 0; i &lt; repeat; i++)
 {
 var ignore = myFunc(R.Next());
 }
 sw.Stop();
 Console.WriteLine("Operation took {0}ms", sw.ElapsedMilliseconds);
}
</code></pre>

Result (10^6 random numbers in positive int32 range):

<pre><code>While loop version:
Operation took 279ms

Linq aggregate:
Operation took 984ms
</code></pre>

blocks|key|2003749|text|乘以-1？请准确地回答你的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003750|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

multiply it by -1? precise your question please...

blocks|key|2277380|text|这应该可以做到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2277381|int+n+=+12345;
int+left+=+n;
int+rev+=+0;
while(Convert.ToBoolean(left))+//+instead+of+left>0+,+to+reverse+signed+numbers+as+well
{
+++int+r+=+left+%25+10;+++
+++rev+=+rev+*+10+%2B+r;
+++left+=+left+/+10;++//left+=+Math.floor(left+/+10);+
}

Console.WriteLine(rev);|code-block|syntax|javascript|2277382|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This should do it:
<pre><code>int n = 12345;
int left = n;
int rev = 0;
while(Convert.ToBoolean(left)) // instead of left&gt;0 , to reverse signed numbers as well
{
 int r = left % 10; 
 rev = rev * 10 + r;
 left = left / 10; //left = Math.floor(left / 10); 
}

Console.WriteLine(rev);
</code></pre>

blocks|key|2001002|text|耶！一条闪闪发光的路。(不，真的。我希望这更像是一个“我该怎么做……”问题，而不是生产中真正需要的东西)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2001003|public+int+Reverse(int+number)+{
++return+int.Parse(number.ToString().Reverse().Aggregate("",+(s,c)+=>+s%2Bc));
}|code-block|syntax|javascript|2001004|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Yay! A bling way. (No, really. I hope that this is more a "How would I do..." question and not something you really need in production)

<pre><code>public int Reverse(int number) {
 return int.Parse(number.ToString().Reverse().Aggregate("", (s,c) =&gt; s+c));
}
</code></pre>

blocks|key|700959|text|你不能。因为计算机在任何情况下都是十六进制的，所以你有必要将数字标记为阿拉伯格式，这在语义上与转换为字符串是相同的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|700960|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can't. Since the computer thinks in hexadecimal in any case, it is necessary for you to tokenise the number into Arabic format, which is semantically identical to the conversion to string.

blocks|key|700589|text|++++///+<summary>
++++///+Reverse+a+int+using+its+sting+representation.
++++///+</summary>
++++private+int+ReverseNumber(int+value)
++++{
++++++++string+textValue+=+value.ToString().TrimStart('-');

++++++++char[]+valueChars+=+textValue.ToCharArray();
++++++++Array.Reverse(valueChars);
++++++++string+reversedValue+=+new+string(valueChars);
++++++++int+reversedInt+=+int.Parse(reversedValue);

++++++++if+(value+<+0)
+++++++++++reversedInt+*=+-1;

++++++++return+reversedInt;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|700590|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> /// &lt;summary&gt;
 /// Reverse a int using its sting representation.
 /// &lt;/summary&gt;
 private int ReverseNumber(int value)
 {
 string textValue = value.ToString().TrimStart('-');

 char[] valueChars = textValue.ToCharArray();
 Array.Reverse(valueChars);
 string reversedValue = new string(valueChars);
 int reversedInt = int.Parse(reversedValue);

 if (value &lt; 0)
 reversedInt *= -1;

 return reversedInt;
 }
</code></pre>

blocks|key|2003842|text|检查下面的简单和容易-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003843|public+int+reverseNumber(int+Number)
{
++int+ReverseNumber+=+0;
++while(Number+>+0)
++{
++++ReverseNumber+=+(ReverseNumber+*+10)+%2B+(Number+%25+10);
++++Number+=+Number+/+10;
++}
++return+ReverseNumber;
}|code-block|syntax|javascript|2003844|参考:+Reverse+number+program+in+c#|offset|length|style|BOLD|2003845|entityMap|0|LINK|mutability|MUTABLE|url|http://www.dotnetbull.com/2013/03/reverse-number-program-in-c.html^0|0|0|0|3|4|S|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@$I|Z|J|10|1|11]]|A|$]]|$1|M|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

Check below simple and easy -

<pre><code>public int reverseNumber(int Number)
{
 int ReverseNumber = 0;
 while(Number &gt; 0)
 {
 ReverseNumber = (ReverseNumber * 10) + (Number % 10);
 Number = Number / 10;
 }
 return ReverseNumber;
}
</code></pre>

Reference : <a href="http://www.dotnetbull.com/2013/03/reverse-number-program-in-c.html" rel="nofollow">Reverse number program in c#</a>

blocks|key|869167|text|旧线程，但我没有看到这个变体：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869168|++++++++int+rev(int+n)+{
++++++++++++string+str+=+new+String(n.ToString().Reverse().ToArray());
++++++++++++return+int.Parse(str);
++++++++}|code-block|syntax|javascript|869169|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Old thread, but I did not see this variation:

<pre><code> int rev(int n) {
 string str = new String(n.ToString().Reverse().ToArray());
 return int.Parse(str);
 }
</code></pre>

blocks|key|869197|text|我看了下面的解决方案。但是当n为负数时呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869198|假设n=+-123456|869199|以下是我为处理负数而添加的一个调整，假设它将以同样的方式威胁负数。|869200|++++++++int+reverse+=+0;
++++++++bool+isNegative+=+false;

++++++++if+(n+<+0)
++++++++++++isNegative+=+true;

++++++++n+=+Math.Abs(n);
++++++++while+(n+>+0)
++++++++{
++++++++++++int+rem+=+n+%25+10;
++++++++++++reverse+=+(reverse+*+10)+%2B+rem;
++++++++++++n+=+n+/+10;
++++++++}

++++++++if+(isNegative)
++++++++++++reverse+=+reverse+*+(-1);

++++++++return+reverse;|code-block|syntax|javascript|869201|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I looked at the following solution. But how about when n is negative?

Lets say n = -123456

Here is a tweak I added to it to handle negative numbers, with that assumption it will threat negative numbs the same way.

<pre><code> int reverse = 0;
 bool isNegative = false;

 if (n &lt; 0)
 isNegative = true;

 n = Math.Abs(n);
 while (n &gt; 0)
 {
 int rem = n % 10;
 reverse = (reverse * 10) + rem;
 n = n / 10;
 }

 if (isNegative)
 reverse = reverse * (-1);

 return reverse;
</code></pre>

blocks|key|701012|text|最短的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|701013|++++++++int+reverse=0;
++++++++int+number+=+21;

++++++++while+(number+>+0)
++++++++{
++++++++++++reverse+=+number+%25+10;
++++++++++++Console.Write(reverse);
++++++++++++number+=+number+/+10;++
++++++++}|code-block|syntax|javascript|701014|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

the shortest solution:

<pre><code> int reverse=0;
 int number = 21;

 while (number &gt; 0)
 {
 reverse = number % 10;
 Console.Write(reverse);
 number = number / 10; 
 }
</code></pre>

blocks|key|2277598|text|int+x+=+9876543;
char[]+arr+=+x.ToString().ToCharArray();
Array.Reverse(arr);
Console.WriteLine(int.Parse(new+string(arr)));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2277599|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int x = 9876543;
char[] arr = x.ToString().ToCharArray();
Array.Reverse(arr);
Console.WriteLine(int.Parse(new string(arr)));
</code></pre>

blocks|key|869299|text|根据Pritam+Karmakar的评论，是的，当给定的数字在末尾有0时，这将不起作用。如果将number作为150传递，它将返回51而不是051。所以我写了代码，按照用户输入的相反顺序显示数字，请检查这一点。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|869300|int+reverseNum+=+0,+reminder,+num;
++++++++Console.WriteLine("Enter+Number+to+Reverse:");
++++++++int.TryParse(Console.ReadLine(),+out+num);
++++++++bool+isZero+=+false;
++++++++int+cnt=0;
++++++++while+(num+>+0)
++++++++{
++++++++++++reminder+=+num+%25+10;
++++++++++++reverseNum+=+(reverseNum+*+10)+%2B+reminder;
++++++++++++num+=+num+/+10;
++++++++++++if+(reverseNum+==+0)
++++++++++++++++isZero+=+true;++++++++++++++++
++++++++++++cnt%2B%2B;
++++++++}
++++++++if+(isZero)
++++++++{
++++++++++++Console.WriteLine(reverseNum.ToString().PadLeft(cnt,+'0'));
++++++++}
++++++++else
++++++++{
++++++++++++Console.WriteLine(reverseNum);
++++++++}
++++++++Console.ReadKey();|code-block|syntax|javascript|869301|entityMap^0|2|G|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

As per Pritam Karmakar comment ,
Yes this will not work when given number has 0 at end.
If you pass number as 150 it will return 51 not 051.
so i had write code to display number in Reverse order as per user input please check this.

<pre><code>int reverseNum = 0, reminder, num;
 Console.WriteLine("Enter Number to Reverse:");
 int.TryParse(Console.ReadLine(), out num);
 bool isZero = false;
 int cnt=0;
 while (num &gt; 0)
 {
 reminder = num % 10;
 reverseNum = (reverseNum * 10) + reminder;
 num = num / 10;
 if (reverseNum == 0)
 isZero = true; 
 cnt++;
 }
 if (isZero)
 {
 Console.WriteLine(reverseNum.ToString().PadLeft(cnt, '0'));
 }
 else
 {
 Console.WriteLine(reverseNum);
 }
 Console.ReadKey();
</code></pre>

blocks|key|869320|text|int+sum+=+0;+int+remainder+=+0;
++++++++int+n+=+123;
++++++++for+(int+i+=+n;+i+>+0;+i+=+i+/+10)
++++++++{
++++++++++++remainder+=+i+%25+10;
++++++++++++sum+=+(sum+*+10)+%2B+remainder;
++++++++}
Console.WriteLine(sum);
Console.ReadLine();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|869321|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int sum = 0; int remainder = 0;
 int n = 123;
 for (int i = n; i &gt; 0; i = i / 10)
 {
 remainder = i % 10;
 sum = (sum * 10) + remainder;
 }
Console.WriteLine(sum);
Console.ReadLine();
</code></pre>

blocks|key|2001164|text|++++public+static+void+Main(string[]+args)
++++{
++++++++int+reversed+=+Reverse(-7360);
++++}

++++public+static+int+Reverse(int+number,+int+left+=+0,+int+right+=+0)
++++{
++++++++if+(number+==+0)+return+left+%2B+right;
++++++++return+Reverse(number+/+10,+left+*+10+%2B+right+*+10,+number+%25+10);
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2001165|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> public static void Main(string[] args)
 {
 int reversed = Reverse(-7360);
 }

 public static int Reverse(int number, int left = 0, int right = 0)
 {
 if (number == 0) return left + right;
 return Reverse(number / 10, left * 10 + right * 10, number % 10);
 }
</code></pre>

blocks|key|869332|text|++++static+void+Main(string[]+args)
++++{
++++++++Console.WriteLine("Please+enter+number");
++++++++int+number+=+Convert.ToInt32(Console.ReadLine());
++++++++Console.WriteLine(ReverseLoop(number));
++++++++Console.WriteLine(ReverseRecursive(number));
++++++++Console.ReadLine();
++++}

++++public+static+int+ReverseRecursive(int+number)
++++{
++++++++int+remainder+=+number+%25+10;
++++++++number+=+number+/+10;
++++++++if+(number+<+1)
++++++++++++return+remainder;
++++++++return+ReverseRecursive(number)+%2B+remainder+*+Convert.ToInt32(Math.Pow(10,+number.ToString().Length));
++++}

++++public+static+int+ReverseLoop(int+number)
++++{
++++++++int+reversed+=+0;
++++++++while+(number+>+0)
++++++++{
++++++++++++reversed+=+reversed+*+10+%2B+(number+%25+10);
++++++++++++number+=+number+/+10;
++++++++}
++++++++return+reversed;
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|869333|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> static void Main(string[] args)
 {
 Console.WriteLine("Please enter number");
 int number = Convert.ToInt32(Console.ReadLine());
 Console.WriteLine(ReverseLoop(number));
 Console.WriteLine(ReverseRecursive(number));
 Console.ReadLine();
 }

 public static int ReverseRecursive(int number)
 {
 int remainder = number % 10;
 number = number / 10;
 if (number &lt; 1)
 return remainder;
 return ReverseRecursive(number) + remainder * Convert.ToInt32(Math.Pow(10, number.ToString().Length));
 }

 public static int ReverseLoop(int number)
 {
 int reversed = 0;
 while (number &gt; 0)
 {
 reversed = reversed * 10 + (number % 10);
 number = number / 10;
 }
 return reversed;
 }
</code></pre>

blocks|key|2003983|text|一个新的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003984|using+System;
using+System.Linq;

public+class+Program
{
++++public+static+void+Main()
++++{
++++++++int+i+=+234;
++++++++int+j+=+-123;
++++++++Console.WriteLine(reverse(i));
++++++++Console.WriteLine(reverse(j));
++++}

++++static+int+reverse(int+i)
++++{
++++++++int+sign+=+i+/+Math.Abs(i);
++++++++var+abs+=+string.Concat(Math.Abs(i).ToString().Reverse());
++++++++return+int.Parse(abs)+*+sign;
++++}
}|code-block|syntax|javascript|2003985|https://dotnetfiddle.net/VGJJFf|offset|length|2003986|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|V|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|T|8|@]|9|@$I|U|J|V|1|W]]|A|$]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|H]]]]

a new one:

<pre><code>using System;
using System.Linq;

public class Program
{
 public static void Main()
 {
 int i = 234;
 int j = -123;
 Console.WriteLine(reverse(i));
 Console.WriteLine(reverse(j));
 }

 static int reverse(int i)
 {
 int sign = i / Math.Abs(i);
 var abs = string.Concat(Math.Abs(i).ToString().Reverse());
 return int.Parse(abs) * sign;
 }
}
</code></pre>

<a href="https://dotnetfiddle.net/VGJJFf" rel="nofollow noreferrer">https://dotnetfiddle.net/VGJJFf</a>

blocks|key|2001198|text|@jerjer和@cfern建议的解决方案是完美的，但它不适用于负整数，如-1234507。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2001199|解决方案：如果数字为负整数，则在调用方法ReverseInt(-1234507)后乘以-1|2001200|++++int+number+=+-1234507;
++++**number+=+number+>+0+?+number+:+(number+*+-1);**

++++int+reverseNumber+=+0;

++++while(number+>+0){
++++++++reverseNumber+=+(reverseNumber+*+10)+%2B++(number+%25+10)+;
++++++++number+=+number+/+10;
++++}
++++Console.WriteLine(reverseNumber);|code-block|syntax|javascript|2001201|entityMap^0|0|7|8|6|W|3|11|8|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

Solution suggested by @jerjer and @cfern is perfect but It will not work for negative int number like -1234507. 

Solution : If number is negative int then multiply by -1 after that call your method ReverseInt(-1234507)

<pre><code> int number = -1234507;
 **number = number &gt; 0 ? number : (number * -1);**

 int reverseNumber = 0;

 while(number &gt; 0){
 reverseNumber = (reverseNumber * 10) + (number % 10) ;
 number = number / 10;
 }
 Console.WriteLine(reverseNumber);
</code></pre>

blocks|key|869399|text|你可以做这样的事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|869400|Convert.ToInt32(new+string(num.ToString().Reverse().ToArray()));|code-block|syntax|javascript|869401|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can do something like this.
<pre><code>Convert.ToInt32(new string(num.ToString().Reverse().ToArray()));
</code></pre>

I came across a question "How can one reverse a number as an integer and not as a string?" Could anyone please help me to find out the answer? Reversal should reverse the decimal digits of the number, i.e. use base 10.

How to reverse a number as an integer and not as a string?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我遇到了一个问题：“如何才能将数字反转为整数而不是字符串？”有人能帮我找出答案吗？反转应该反转数字的小数位数，即使用基数10。

问如何将数字转换为整数而不是字符串？
EN

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问如何将数字转换为整数而不是字符串？EN