C#中定义数组和一个简单排序算法实现。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApp20110730
{
class Program
{
static void Main(string[] args)
{
int n = 18;
int[] a = new int[18]{ 0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5};
int[] b = new int[18] { 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5 };

Console.WriteLine("排序前:");
Console.WriteLine("初始化输入A【18】：");
foreach (var item in a)
{
Console.Write(item);
}
Console.WriteLine("");
Console.WriteLine("初始化输入A【18】结束。");

Console.WriteLine("初始化输出B【18】：");
foreach (var item in b)
{
Console.Write(item);
}
Console.WriteLine("");
Console.WriteLine("初始化输入B【18】结束。");

Console.WriteLine("开始排序。。。。");
Sort(n,a,b);
Console.WriteLine("排序结束。");
Console.WriteLine("排序后:");
Console.WriteLine("初始化输入A【18】：");
foreach (var item in a)
{
Console.Write(item);
}
Console.WriteLine("");
Console.WriteLine("初始化输入A【18】结束。");

Console.WriteLine("初始化输出B【18】：");
foreach (var item in b)
{
Console.Write(item);
}
Console.WriteLine("");
Console.WriteLine("初始化输入B【18】结束。");
}

private static void Sort(int n, int[] a, int[] b)
{
try
{
int r = 6;
int[] R = new int[r];
for (int i = 0; i < r; i++)
{
R[i] = 0;
}

for (int i = 0; i < n; i++)
{
R[a[i]] = R[a[i]] + 1;
}

for (int i = 1; i < r; i++)
{
R[i] = R[i] + R[i - 1];
}

for (int i =(n-1); i >= 0; i--)
{
b[R[a[i]] - 1] = a[i];
R[a[i]] = R[a[i]] - 1;
}
}
catch (Exception ex)
{
throw new Exception(ex.Message);
}
}
}
}

　　

C#中定义数组

一、一维：
int[] numbers = new int[]{1,2,3,4,5,6}; //不定长
int[] numbers = new int[3]{1,2,3};//定长
二、多维
int[,] numbers = new int[,]{{1,2,3},{1,2,3}}; //不定长
int[,] numbers = new int[2,2]{{1,2},{1,2}}; //定长

三、例子
A：int[] mf1=new int[6];
//注意初始化数组的范围,或者指定初值; //包含6个元素的一维整数数组，初值1，2，3，4，5，6
int[] mf2=new int[6]{1,2,3,4,5,6};

B：//一维字符串数组,如果提供了初始值设定项，则还可以省略 new 运算符
string[] mf3={"c","c++","c#"};

C：//一维对象数组
Object[] mf4 = new Object[5] { 26, 27, 28, 29, 30 };

D：//二维整数数组，初值mf5[0,0]=1,mf5[0,1]=2,mf5[1,0]=3,mf5[1,1]=4
int[,] mf5=new int[,]{{1,2},{3,4}};

E：//6*6的二维整型数组
int[,] mf6=new mf[6,6];

四、取得数组元素个数：
int b;
b = sizeof (a)/sizeof (*a);