C# % (mod) 解释
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% (mod) explanation
提问by Wolfy
Today I was writing a program in C#, and I used %to calculate some index... My program didn't work, so I debugged it and I realized that "%" is not working like in other program languages that I know.
今天我正在用 C# 编写一个程序，我用%来计算一些索引......我的程序不起作用，所以我调试了它，我意识到“ %”不像我知道的其他程序语言那样工作。
For example:
例如：
In Python %returns values like this:
在 Python 中%返回这样的值：
for x in xrange (-5, 6):
print x, "% 5 =", x % 5
-5 % 5 = 0
-4 % 5 = 1
-3 % 5 = 2
-2 % 5 = 3
-1 % 5 = 4
0 % 5 = 0
1 % 5 = 1
2 % 5 = 2
3 % 5 = 3
4 % 5 = 4
5 % 5 = 0
In C#:
在 C# 中：
for (int i = -5; i < 6; i++)
{
Console.WriteLine(i + " % 5 = " + i % 5);
}
-5 % 5 = 0
-4 % 5 = -4
-3 % 5 = -3
-2 % 5 = -2
-1 % 5 = -1
0 % 5 = 0
1 % 5 = 1
2 % 5 = 2
3 % 5 = 3
4 % 5 = 4
5 % 5 = 0
Did I do something wrong or is %not working like it should?
我做错了什么还是%没有像它应该的那样工作？
采纳答案by David Heffernan
As explained in the comments, the different behaviour is by design. The different languages just ascribe different meanings to the %operator.
正如评论中所解释的，不同的行为是设计使然。不同的语言只是赋予%运算符不同的含义。
You ask:
你问：
How can I use modulus operator in C#?
如何在 C# 中使用模数运算符？
You can define a modulus operator yourself that behaves the same way as the Python %operator:
您可以自己定义一个模运算符，其行为方式与 Python%运算符相同：
int mod(int a, int n)
{
int result = a % n;
if ((result<0 && n>0) || (result>0 && n<0)) {
result += n;
}
return result;
}
回答by Hauleth
Check table on the right side: http://en.wikipedia.org/wiki/Modulo_operation
回答by Niet the Dark Absol
Both answers are correct. Although personally I think the "always positive" one makes more sense.
两个答案都是正确的。虽然我个人认为“总是积极的”更有意义。
You can define your own modulus function that only gives positive answers like this:
您可以定义自己的模函数，它只给出如下肯定的答案：
int mod(int a, int n) {
return ((a%n)+n) % n;
}
回答by Sklivvz
In modular arithmetic, one defines classesof numbers based on the modulo. In other words, in modulo marithmetic, a number nis equivalent (read: the same) to n+ m, n- m, n+ 2m, n- 2m, etc.
在模运算，一个定义的类号的基于模。换句话说，在模m算术中，数字n等价于（读作：相同）到n+ m、n- m、n+ 2m、n- 2m等。
One defines m"baskets" and every number falls in one (and only one) of them.
一个定义了m 个“篮子”，每个数字都落在其中一个（并且只有一个）中。
Example: one can say "It's 4:30 pm" or one can say "It's 16:30". Both forms mean exactly the same time, but are different representationsof it.
例如：可以说“现在是下午 4:30”或可以说“现在是 16:30”。两种形式都意味着完全相同的时间，但对时间的表示不同。
Thus both, the Python and C# results are correct! The numbers are the samein the modulo 5arithmetic you chose. It would also have been mathematicallycorrect to return (5, 6, 7, 8, 9) for example. Just a bit odd.
因此，Python 和 C# 结果都是正确的！您选择的模5算术中的数字相同。例如，返回 (5, 6, 7, 8, 9)在数学上也是正确的。只是有点奇怪。
As for the choice of representation (in other words, the choice on how to represent negative numbers), that is just a case of different design choices between the two languages.
至于表示的选择（换言之，如何表示负数的选择），这只是两种语言之间设计选择不同的情况。
However, that is not at all what the % operator actually does in C#. The % operator is not the canonical modulus operator; it is the remainder operator. The A % B operator actually answer the question "If I divided A by B using integer arithmetic, what would the remainder be?"
但是，这完全不是 % 运算符在 C# 中实际执行的操作。% 运算符不是规范模数运算符；它是余数运算符。A % B 运算符实际上回答了“如果我使用整数算法将 A 除以 B，余数是多少？”的问题。
— What's the difference? Remainder vs Modulusby Eric Lippert
Quick snippet to get the canonical modulus:
获取规范模数的快速片段：
return ((n % m) + m) % m;
Test implementation:
测试实现：
Mono/C#:
单声道/C#：
machine:~ user$ cat mod.cs
using System;
public class Program
{
public static void Main (string[] args)
{
Console.WriteLine(Mod(-2, 5));
Console.WriteLine(Mod(-5, 5));
Console.WriteLine(Mod(-2, -5));
}
public static int Mod (int n, int m)
{
return ((n % m) + m) % m;
}
}
machine:~ user$ mono mod.exe
3
0
-2
Python:
蟒蛇：
machine:~ user$ cat mod.py
print -2%5;
print -5%5;
print -2%-5;
machine:~ user$ python mod.py
3
0
-2
