blocks|key|1297954|text|请注意，C#和C%2B%2B的%25运算符实际上不是模数，而是余数。在你的例子中，你需要的模数公式是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1297955|float+nfmod(float+a,float+b)
{
++++return+a+-+b+*+floor(a+/+b);
}|code-block|syntax|javascript|1297956|你必须用C#+(或C%2B%2B)重新编码，但这是你得到模数而不是余数的方法。|1297957|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Please note that C# and C++'s % operator is actually NOT a modulo, it's remainder. The formula for modulo that you want, in your case, is:

<pre><code>float nfmod(float a,float b)
{
 return a - b * floor(a / b);
}
</code></pre>

You have to recode this in C# (or C++) but this is the way you get modulo and not a remainder.

blocks|key|1298014|text|只使用一次%25的单行实现：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1298015|int+mod(int+k,+int+n)+{++return+((k+%25=+n)+<+0)+?+k%2Bn+:+k;++}|code-block|syntax|javascript|1298016|entityMap^0|5|1|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|$]]

Single-line implementation using <code>%</code> only once:

<pre><code>int mod(int k, int n) { return ((k %= n) &lt; 0) ? k+n : k; }
</code></pre>

blocks|key|1298103|text|比较两个主要答案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1298104|(x%25m+%2B+m)%25m;|code-block|syntax|javascript|1298105|和|1298106|int+r+=+x%25m;
return+r<0+?+r%2Bm+:+r;|1298107|没有人真正提到第一个答案可能会抛出OverflowException，而第二个答案不会。更糟糕的是，在默认未检查上下文的情况下，第一个答案可能会返回错误的答案(例如参见mod(int.MaxValue+-+1,+int.MaxValue)+)。因此，第二个答案不仅看起来更快，而且更正确。|offset|length|style|CODE|1298108|entityMap^0|0|0|0|0|H|H|2C|Z|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Comparing two predominant answers

<pre><code>(x%m + m)%m;
</code></pre>

and

<pre><code>int r = x%m;
return r&lt;0 ? r+m : r;
</code></pre>

Nobody actually mentioned the fact that the first one may throw an <code>OverflowException</code> while the second one won't. Even worse, with default unchecked context, the first answer may return the wrong answer (see <code>mod(int.MaxValue - 1, int.MaxValue)</code> for example). So the second answer not only seems to be faster, but also more correct.

blocks|key|1512620|text|只需将您的模数(arrayLength)与%25的负值相加，就可以了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1512621|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|$]]

Just add your modulus (arrayLength) to the negative result of % and you'll be fine.

blocks|key|1512704|text|添加一些理解。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1512705|通过Euclidean+definition，mod结果必须始终是正的。|offset|length|1512706|例如：|1512707|+int+n+=+5;
+int+x+=+-3;

+int+mod(int+n,+int+x)
+{
+++++return+((n%25x)%2Bx)%25x;
+}|code-block|syntax|javascript|1512708|输出：|1512709|+-1|1512710|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Modulo_operation^0|0|2|K|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@$D|10|E|11|1|12]]|A|$]]|$1|F|3|G|5|6|7|13|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

Adding some understanding.

By <a href="https://en.wikipedia.org/wiki/Modulo_operation" rel="nofollow">Euclidean definition</a> the mod result must be always positive.

Ex:

<pre><code> int n = 5;
 int x = -3;

 int mod(int n, int x)
 {
 return ((n%x)+x)%x;
 }
</code></pre>

Output:

<pre><code> -1
</code></pre>

blocks|key|1298037|text|对于更具性能意识的开发人员|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1298038|uint+wrap(int+k,+int+n)+((uint)k)%25n|code-block|syntax|javascript|1298039|一个小的性能比较|1298040|Modulo:+00:00:07.2661827+((n%25x)%2Bx)%25x)
Cast:+++00:00:03.2202334+((uint)k)%25n
If:+++++00:00:13.5378989+((k+%25=+n)+<+0)+?+k%2Bn+:+k|1298041|至于cast+to+uint的性能成本，请看here|offset|length|1298042|entityMap|0|LINK|mutability|MUTABLE|url|http://www.codeproject.com/Articles/8052/Type-casting-impact-over-execution-performance-in^0|0|0|0|0|M|4|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

For the more performance aware devs

<pre><code>uint wrap(int k, int n) ((uint)k)%n
</code></pre>

A small performance comparison

<pre><code>Modulo: 00:00:07.2661827 ((n%x)+x)%x)
Cast: 00:00:03.2202334 ((uint)k)%n
If: 00:00:13.5378989 ((k %= n) &lt; 0) ? k+n : k
</code></pre>

As for performance cost of cast to uint have a look <a href="http://www.codeproject.com/Articles/8052/Type-casting-impact-over-execution-performance-in" rel="nofollow noreferrer">here</a>

blocks|key|1512862|text|如果你的除数是正的，这里的所有答案都能很好地工作，但它不是完全的。下面是我的实现，它总是在[0,+b)的范围内返回，使得输出的符号与除数的符号相同，允许负的除数作为输出范围的端点。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1512863|PosMod(5,+3)返回2|1512864|PosMod(-5,+3)返回1|1512865|PosMod(5,+-3)返回-1|1512866|PosMod(-5,+-3)返回-2|1512867|++++///+<summary>
++++///+Performs+a+canonical+Modulus+operation,+where+the+output+is+on+the+range+[0,+b).
++++///+</summary>
++++public+static+real_t+PosMod(real_t+a,+real_t+b)
++++{
++++++++real_t+c+=+a+%25+b;
++++++++if+((c+<+0+&&+b+>+0)+%7C%7C+(c+>+0+&&+b+<+0))+
++++++++{
++++++++++++c+%2B=+b;
++++++++}
++++++++return+c;
++++}|code-block|syntax|javascript|1512868|(其中real_t可以是任何数字类型)|1512869|entityMap^0|19|6|0|0|C|E|1|0|0|D|F|1|0|0|D|F|2|0|0|E|G|2|0|0|3|6|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1E|8|@$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]]|D|@]|E|$]]|$1|N|3|O|5|P|7|1J|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|1K|8|@$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|V|$]]

All of the answers here work great if your divisor is positive, but it's not quite complete. Here is my implementation which always returns on a range of <code>[0, b)</code>, such that the sign of the output is the same as the sign of the divisor, allowing for negative divisors as the endpoint for the output range.

<code>PosMod(5, 3)</code> returns <code>2</code> 
<code>PosMod(-5, 3)</code> returns <code>1</code> 
<code>PosMod(5, -3)</code> returns <code>-1</code> 
<code>PosMod(-5, -3)</code> returns <code>-2</code>

<pre><code> /// &lt;summary&gt;
 /// Performs a canonical Modulus operation, where the output is on the range [0, b).
 /// &lt;/summary&gt;
 public static real_t PosMod(real_t a, real_t b)
 {
 real_t c = a % b;
 if ((c &lt; 0 &amp;&amp; b &gt; 0) || (c &gt; 0 &amp;&amp; b &lt; 0)) 
 {
 c += b;
 }
 return c;
 }
</code></pre>

(where <code>real_t</code> can be any number type)

blocks|key|1298161|text|这是我的一个基于this+answer的正整数线性程序|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1298162|用法：|1298163|(-7).Mod(3);+//+returns+2|code-block|syntax|javascript|1298164|实施：|1298165|static+int+Mod(this+int+a,+int+n)+=>+(((a+%25=+n)+<+0)+?+n+:+0)+%2B+a;|1298166|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/23214321/7032856^0|8|B|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|6|7|10|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|13|8|@]|9|@]|C|$I|J]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

Here's my one liner for positive integers, based on <a href="https://stackoverflow.com/a/23214321/7032856">this answer</a>:
usage:
<pre><code>(-7).Mod(3); // returns 2
</code></pre>
implementation:
<pre><code>static int Mod(this int a, int n) =&gt; (((a %= n) &lt; 0) ? n : 0) + a;
</code></pre>

blocks|key|1512932|text|mod函数有很多实现，我认为有必要列出所有这些实现-至少根据Wikipedia的说法，我相信还有更多。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1512933|//+Important+to+be+able+to+use+`MathF`.
using+System;

public+static+class+MathFUtils+{
++++public+static+class+Mod+{
++++++++public+static+float+Trunc(float+a,+float+b)+=>
++++++++++++a+-+b+*+((int)(a+/+b));

++++++++public+static+float+Round(float+a,+float+b)+=>
++++++++++++a+-+b+*+MathF.Round(a+/+b);

++++++++public+static+float+Floor(float+a,+float+b)+=>
++++++++++++a+-+b+*+MathF.Floor(a+/+b);

++++++++public+static+float+Ceil(float+a,+float+b)+=>
++++++++++++a+-+b+*+MathF.Ceiling(a+/+b);

++++++++public+static+float+Euclidean(float+a,+float+b)+=>
++++++++++++a+-+MathF.Abs(b)+*+MathF.Floor(a+/+MathF.Abs(b));
++++}
}|code-block|syntax|javascript|1512934|1512935|根据维基百科(以及我的经验)，坚持使用Euclidean。在数学和概率属性方面，它是最有用的。如果你需要Trunc，那么我相信%25就是这样做的。|1512936|此外，对于那些可能对它们各自的功能和方式感到困惑的人，我强烈建议您阅读维基百科的文章(即使很难)，并查看每个表示的图像。|1512937|当然，这些不一定是性能最好的，但它们确实起作用了。如果你关心性能，我建议你找一个当地的C#之神，或者在他们通过我们的凡人层面时去问一个。|1512938|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Modulo_operation^0|0|3|U|9|0|0|0|0|J|9|1G|5|1R|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]]|E|$]]|$1|F|3|G|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|16|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|1E|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1F|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]]]

There are many implementations of the <code>mod</code> function, and I think it's worth it to list all of them --- at least according to <a href="https://en.wikipedia.org/wiki/Modulo_operation" rel="nofollow noreferrer">Wikipedia</a>, I'm sure there are more.
<pre class="lang-cs prettyprint-override"><code>// Important to be able to use `MathF`.
using System;

public static class MathFUtils {
 public static class Mod {
 public static float Trunc(float a, float b) =&gt;
 a - b * ((int)(a / b));

 public static float Round(float a, float b) =&gt;
 a - b * MathF.Round(a / b);

 public static float Floor(float a, float b) =&gt;
 a - b * MathF.Floor(a / b);

 public static float Ceil(float a, float b) =&gt;
 a - b * MathF.Ceiling(a / b);

 public static float Euclidean(float a, float b) =&gt;
 a - MathF.Abs(b) * MathF.Floor(a / MathF.Abs(b));
 }
}
</code></pre>
<hr />
According to the Wikipedia (as well as my experience) stick to <code>Euclidean</code>. It is the most useful in terms of mathematical and probabilistic properties. If you ever need <code>Trunc</code>, then I believe <code>%</code> does just that.
Also, for those of you who might be confused as to what each of them do, and how, I highly recommend reading the Wikipedia article (even if it's hard) and looking at the images of each representation.
Of course these are not necessarily the most performant, but they do work. If you are concerned about performance, I recommend finding a local C# god, or asking one as they pass through our mortal plane.

I'm trying to mod an integer to get an array position so that it will loop round. Doing <code>i % 
arrayLength</code> works fine for positive numbers but for negative numbers it all goes wrong.

<pre><code> 4 % 3 == 1
 3 % 3 == 0
 2 % 3 == 2
 1 % 3 == 1
 0 % 3 == 0
-1 % 3 == -1
-2 % 3 == -2
-3 % 3 == 0
-4 % 3 == -1
</code></pre>

so i need an implementation of

<pre><code>int GetArrayIndex(int i, int arrayLength)
</code></pre>

such that

<pre><code>GetArrayIndex( 4, 3) == 1
GetArrayIndex( 3, 3) == 0
GetArrayIndex( 2, 3) == 2
GetArrayIndex( 1, 3) == 1
GetArrayIndex( 0, 3) == 0
GetArrayIndex(-1, 3) == 2
GetArrayIndex(-2, 3) == 1
GetArrayIndex(-3, 3) == 0
GetArrayIndex(-4, 3) == 2
</code></pre>

I've done this before but for some reason it's melting my brain today :(

Mod of negative number is melting my brain

我试图对一个整数进行mod，以获得一个数组位置，这样它就会循环。对于正数来说，执行i %  arrayLength工作得很好，但是对于负数，这一切都是错误的。 4 % 3 == 1 3 % 3 == 0 2 % 3 == 2 1 % 3 == 1 0 % 3 == 0-1 % 3 == -1-2 % 3 == -2-3 % 3 == 0-4 % 3 == -1所以我需要一个实现int GetArr

问负数的模式正在融化我的大脑
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