blocks|key|1694656|text|使用Math.Ceiling始终向上舍入。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1694657|int+number+=+236;
number+=+(int)(Math.Ceiling(number+/+10.0d)+*+10);|code-block|syntax|javascript|1694658|模数(%25)得到余数，所以你得到：|1694659|//+number+=+236+%2B+10+-+6|1694660|将其放入一个扩展方法中|1694661|public+static+int+roundupbyten(this+int+i){
++++//+return+i+%2B+(10+-+i+%25+10);+<--+logic+error.+Oops!
++++return+(int)(Math.Ceiling(i+/+10.0d)*10);+//+fixed
}

//+call+like+so:
int+number+=+236.roundupbyten();|1694662|上图编辑:我应该凭直觉使用Math.Ceiling|1694663|I+blogged+about+this+when+calculating+UPC+check+digits。|offset|length|1694664|entityMap|0|LINK|mutability|MUTABLE|url|http://atomiton.blogspot.com/2007/12/calculating-checkdigit-using.html^0|0|0|0|0|0|0|0|2|1G|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@$S|1A|T|1B|1|1C]]|A|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

Use Math.Ceiling to always round up.
<pre><code>int number = 236;
number = (int)(Math.Ceiling(number / 10.0d) * 10);
</code></pre>
Modulus(%) gets the remainder, so you get:
<pre><code>// number = 236 + 10 - 6
</code></pre>
Put that into an extension method
<pre><code>public static int roundupbyten(this int i){
 // return i + (10 - i % 10); &lt;-- logic error. Oops!
 return (int)(Math.Ceiling(i / 10.0d)*10); // fixed
}

// call like so:
int number = 236.roundupbyten();
</code></pre>
above edited: I should've gone with my first instinct to use Math.Ceiling
I <a href="http://atomiton.blogspot.com/2007/12/calculating-checkdigit-using.html" rel="noreferrer">blogged about this when calculating UPC check digits</a>.

blocks|key|1479469|text|在类库中没有内置的函数可以做到这一点。最接近的是System.Math.Round()，它只用于将Decimal和Double类型的数字舍入到最接近的整数值。但是，如果您使用的是.NET+3.5，则可以将语句包装在扩展方法中，这将允许您更干净地使用该函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1479470|public+static+class+ExtensionMethods
{
++++public+static+int+RoundOff+(this+int+i)
++++{
++++++++return+((int)Math.Round(i+/+10.0))+*+10;
++++}
}

int+roundedNumber+=+236.RoundOff();+//+returns+240
int+roundedNumber2+=+11.RoundOff();+//+returns+10|code-block|syntax|javascript|1479471|如果您正在使用旧版本的.NET框架进行编程，只需从RoundOff函数中删除"this“，并像这样调用该函数：|1479472|int+roundedNumber+=+ExtensionMethods.RoundOff(236);+//+returns+240
int+roundedNumber2+=+ExtensionMethods.RoundOff(11);+//+returns+10|1479473|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/system.math.round.aspx^0|O|J|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

There is no built-in function in the class library that will do this. The closest is <a href="http://msdn.microsoft.com/en-us/library/system.math.round.aspx" rel="noreferrer">System.Math.Round()</a> which is only for rounding numbers of types Decimal and Double to the nearest integer value. However, you can wrap your statement up in a extension method, if you are working with .NET 3.5, which will allow you to use the function much more cleanly.

<pre><code>public static class ExtensionMethods
{
 public static int RoundOff (this int i)
 {
 return ((int)Math.Round(i / 10.0)) * 10;
 }
}

int roundedNumber = 236.RoundOff(); // returns 240
int roundedNumber2 = 11.RoundOff(); // returns 10
</code></pre>

If you are programming against an older version of the .NET framework, just remove the "this" from the RoundOff function, and call the function like so:

<pre><code>int roundedNumber = ExtensionMethods.RoundOff(236); // returns 240
int roundedNumber2 = ExtensionMethods.RoundOff(11); // returns 10
</code></pre>

blocks|key|1479501|text|这可能有点晚了，但我想这可能有一天会有很好的帮助……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1479502|我已经尝试过了：|1479503|public+int+RoundOff(int+number,+int+interval){
++++int+remainder+=+number+%25+interval;
++++number+%2B=+(remainder+<+interval+/+2)+?+-remainder+:+(interval+-+remainder);
++++return+number;
}|code-block|syntax|javascript|1479504|要使用以下命令：|1479505|int+number+=+11;
int+roundednumber+=+RoundOff(number,+10);|1479506|这样，您可以选择是向上舍入还是向下舍入间隔的一半。=)|1479507|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This might be a little too late but I guess this might be of good help someday...

I have tried this:

<pre><code>public int RoundOff(int number, int interval){
 int remainder = number % interval;
 number += (remainder &lt; interval / 2) ? -remainder : (interval - remainder);
 return number;
}
</code></pre>

To use:

<pre><code>int number = 11;
int roundednumber = RoundOff(number, 10);
</code></pre>

This way, you have the option whether if the half of the interval will be rounded up or rounded down. =)

blocks|key|1479571|text|我不喜欢引入Math库，也不喜欢使用浮点数，所以我的建议是只做整数运算，就像下面这样，我向上舍入到下一个1K。如果不想重复，可以将其包装在方法或lambda代码片段中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1479572|int+MyRoundedUp1024Int+=+((lSomeInteger+%2B+1023)+/+1024)+*+1024;|code-block|syntax|javascript|1479573|我没有运行性能测试与其他方法相比，但我敢打赌，这是最快的方式来做这一点，也许是移位和旋转的比特版本。|1479574|entityMap^0|6|4|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

I prefer to not bring in the <code>Math</code> library nor go to floating point so my suggestion is just do integer arithmetic like below where I round up to the next 1K. Wrap it in a method or lambda snippet or something if you don't want to repeat.

<pre><code>int MyRoundedUp1024Int = ((lSomeInteger + 1023) / 1024) * 1024;
</code></pre>

I have not run performance tests on this vs. other the ways but I'd bet it is the fastest way to do this save maybe a shifting and rotating of bits version of this.

blocks|key|1479644|text|老问题，但这里有一种方法来做已经被要求的事情，加上我扩展了它，能够将任何数字舍入到您想要的符号无花果的数量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1479645|++++private+double+Rounding(double+d,+int+digits)
++++{
++++++++int+neg+=+1;
++++++++if+(d+<+0)
++++++++{
++++++++++++d+=+d+*+(-1);
++++++++++++neg+=+-1;
++++++++}

++++++++int+n+=+0;
++++++++if+(d+>+1)
++++++++{
++++++++++++while+(d+>+1)
++++++++++++{
++++++++++++++++d+=+d+/+10;
++++++++++++++++n%2B%2B;
++++++++++++}
++++++++++++d+=+Math.Round(d+*+Math.Pow(10,+digits));
++++++++++++d+=+d+*+Math.Pow(10,+n+-+digits);
++++++++}
++++++++else
++++++++{
++++++++++++while+(d+<+0.1)
++++++++++++{
++++++++++++++++d+=+d+*+10;
++++++++++++++++n%2B%2B;
++++++++++++}
++++++++++++d+=+Math.Round(d+*+Math.Pow(10,+digits));
++++++++++++d+=+d+/+Math.Pow(10,+n+%2B+digits);
++++++++}

++++++++return+d*neg;
++++}


+++private+void+testing()
+++{
+++++++double+a+=+Rounding(1230435.34553,3);
+++++++double+b+=+Rounding(0.004567023523,4);
+++++++double+c+=+Rounding(-89032.5325,2);
+++++++double+d+=+Rounding(-0.123409,4);
+++++++double+e+=+Rounding(0.503522,1);
+++++++Console.Write(a.ToString()+%2B+"\n"+%2B+b.ToString()+%2B+"\n"+%2B+
+++++++++++c.ToString()+%2B+"\n"+%2B+d.ToString()+%2B+"\n"+%2B+e.ToString()+%2B+"\n");
+++}|code-block|syntax|javascript|1479646|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 question but here is a way to do what has been asked plus I extended it to be able to round any number to the number of sig figs you want.

<pre><code> private double Rounding(double d, int digits)
 {
 int neg = 1;
 if (d &lt; 0)
 {
 d = d * (-1);
 neg = -1;
 }

 int n = 0;
 if (d &gt; 1)
 {
 while (d &gt; 1)
 {
 d = d / 10;
 n++;
 }
 d = Math.Round(d * Math.Pow(10, digits));
 d = d * Math.Pow(10, n - digits);
 }
 else
 {
 while (d &lt; 0.1)
 {
 d = d * 10;
 n++;
 }
 d = Math.Round(d * Math.Pow(10, digits));
 d = d / Math.Pow(10, n + digits);
 }

 return d*neg;
 }


 private void testing()
 {
 double a = Rounding(1230435.34553,3);
 double b = Rounding(0.004567023523,4);
 double c = Rounding(-89032.5325,2);
 double d = Rounding(-0.123409,4);
 double e = Rounding(0.503522,1);
 Console.Write(a.ToString() + "\n" + b.ToString() + "\n" + 
 c.ToString() + "\n" + d.ToString() + "\n" + e.ToString() + "\n");
 }
</code></pre>

How to, in C# round any value to 10 interval? For example, if I have 11, I want it to return 10, if I have 136, then I want it to return 140. 

I can easily do it by hand

<pre><code>return ((int)(number / 10)) * 10;
</code></pre>

But I am looking for an builtin algorithm to do this job, something like Math.Round(). The reason why I won't want to do by hand is that I don't want to write same or similar piece of code all over my projects, even for something as simple as the above.

Built in .Net algorithm to round value to the nearest 10 interval

性能测试

如何，在C#中将任意值舍入为10的区间？例如，如果我有11，我希望它返回10，如果我有136，那么我希望它返回140。我可以很容易地手工做这件事return ((int)(number / 10)) * 10;但我正在寻找一种内置算法来完成这项工作，比如Math.Round()。我不想手动完成的原因是，我不想在我的项目...

问内置.Net算法，可将数值舍入到最接近的10个区间
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