操作系统PV操作——进程同步问题，C#实现
在C#中，用于同步的主要是Mutex类与Semaphore类。位于System.Threading命名空间中。
在这两个种对象的方法中，P操作对应的是WaitOne()方法，V操作对应的是ReleaseMutex()与Release()方法。
下面是用C#解决几大PV操作经典问题及其变形的代码。
一、生产者消费者问题
1.最简单的情况：一个生产者，一个消费者，共用一个缓冲区进行生产消费。
1 using System;
2 using System.Threading;
3
4 namespace ProducerCustomer1_SingleBuffer
5 {
6 class ProducerCustomer1_SingleBuffer
7 {
8 static Mutex mutex = new Mutex();
9 static void Main(string[] args)
10 {
11 new Thread(new ThreadStart(Producer)).Start();
12 new Thread(new ThreadStart(Customer)).Start();
13 Console.Read();
14 }
15
16 private static void Producer()
17 {
18 while (true)
19 {
20 mutex.WaitOne();
21 Console.WriteLine("Producer is working!");
22 mutex.ReleaseMutex();
23 Thread.Sleep(400);
24 }
25 }
26
27 private static void Customer()
28 {
29 while (true)
30 {
31 mutex.WaitOne();
32 Console.WriteLine("customer is working");
33 mutex.ReleaseMutex();
34 Thread.Sleep(400);
35 }
36 }
37 }
38 }2.现在稍微复杂一些，生产者与消费者共用一个大小为n的环形缓冲区。本例中n取10.
1 using System;
2 using System.Threading;
3
4 namespace ProducerCustomer2_MultiBuffer
5 {
6 class ProducerCustomer2_MultiBuffer
7 {
8 static Mutex mutex = new Mutex();
9 static Semaphore empty = new Semaphore(10,10);
10 static Semaphore full = new Semaphore(0,10);
11 static int[] buffer = new int[10];
12 static Random rand = new Random();
13
14 static void Main(string[] args)
15 {
16 new Thread(new ThreadStart(Producer)).Start();
17 new Thread(new ThreadStart(Customer)).Start();
18 Console.Read();
19 }
20
21 private static void Producer()
22 {
23 uint ppointer=0;
24 int temp;
25 while (true)
26 {
27 //!!!!Attention！！！！！NEVER MIX Orders
28 //如果empty与mutex的顺序反了，就会发生死锁！
29 empty.WaitOne();
30 mutex.WaitOne();
31 temp = rand.Next(1, 100);
32 buffer[ppointer] = temp;
33 Console.WriteLine("Producer works at {0} with {1}",ppointer,temp);
34 ppointer = (ppointer + 1)%10;
35 mutex.ReleaseMutex();
36 full.Release();
37 Thread.Sleep(400);
38 }
39 }
40
41 private static void Customer()
42 {
43 uint cpointer=0;
44 int temp;
45 while (true)
46 {
47 full.WaitOne();
48 mutex.WaitOne();
49 temp = rand.Next(1, 100);
50 temp = buffer[cpointer];
51 Console.WriteLine("Customer gains at {0} with {1}", cpointer, temp);
52 cpointer = (cpointer + 1) % 10;
53 mutex.ReleaseMutex();
54 empty.Release();
55 Thread.Sleep(400);
56 }
57 }
58 }
59 }3.问题再升级：
桌上有一个空盘子，只允许放一个水果。爸爸可以放苹果。也可以放橘子。儿子要吃苹果，女儿要吃橘子。试实现之。
1 using System;
2 using System.Threading;
3
4 namespace ProducerCustomer3_Fruit
5 {
6 internal class ProducerCustomer3_Fruit
7 {
8 private static Mutex mutex = new Mutex();
9 private static Semaphore Sempty = new Semaphore(1, 1);
10 private static Semaphore Sorange = new Semaphore(0, 1);
11 private static Semaphore Sapple = new Semaphore(0, 1);
12
13 private static Fruit buffer = 0;
14 private static Random rand = new Random();
15
16 public enum Fruit
17 {
18 Empty,
19 Apple,
20 Orange
21 };
22
23 private static void Main(string[] args)
24 {
25 new Thread(new ThreadStart(Papa)).Start();
26 new Thread(new ThreadStart(Son)).Start();
27 new Thread(new ThreadStart(Daughter)).Start();
28 Console.Read();
29 }
30
31 private static void Papa()
32 {
33 while (true)
34 {
35 int temp = 0;
36 Sempty.WaitOne();
37 mutex.WaitOne();
38 temp = rand.Next(1,3);
39 if (temp == 1)
40 {
41 Console.WriteLine("Papa put an apple");
42 buffer = Fruit.Apple;
43 mutex.ReleaseMutex();
44 Sapple.Release();
45 }
46 else
47 {
48 Console.WriteLine("Papa put an Orange");
49 buffer = Fruit.Orange;
50 mutex.ReleaseMutex();
51 Sorange.Release();
52 }
53 Thread.Sleep(400);
54 }
55 }
56
57 private static void Son()
58 {
59 while (true)
60 {
61 Sapple.WaitOne();
62 mutex.WaitOne();
63 Console.WriteLine("Son eat an {0}", buffer);
64 mutex.ReleaseMutex();
65 Sempty.Release();
66 Thread.Sleep(400);
67 }
68 }
69
70 private static void Daughter()
71 {
72 while (true)
73 {
74 Sorange.WaitOne();
75 mutex.WaitOne();
76 Console.WriteLine("Daughter eat an {0}", buffer);
77 mutex.ReleaseMutex();
78 Sempty.Release();
79 Thread.Sleep(400);
80 }
81 }
82 }
83 }二、哲学家进餐问题
五个哲学家围着桌子共同进餐，每个哲学家两侧各有一支筷子。试设计同步算法，使哲学家都能吃上饭。
1 using System;
2 using System.Threading;
3
4 namespace PhilosophersProblem
5 {
6 class PhilosophersProblem
7 {
8 static Semaphore[] chopsticks = { new Semaphore(1, 1)
9 , new Semaphore(1, 1)
10 , new Semaphore(1, 1)
11 , new Semaphore(1, 1),
12 new Semaphore(1, 1), };
13
14 static void Main(string[] args)
15 {
16 for (int i = 0; i < 5; i++)
17 {
18 new Thread(new ParameterizedThreadStart(philosopher)).Start(i);
19 }
20 Console.Read();
21 }
22
23 private static void philosopher(object input)
24 {
25 int i = (int)input;
26 chopsticks[i].WaitOne();
27 chopsticks[(i + 1)%5].WaitOne();
28
29 Console.WriteLine("Philosopher{0}: I am eating",i);
30 Thread.Sleep(1000);
31
32 chopsticks[i].Release();
33 chopsticks[(i + 1)%5].Release();
34 }
35 }
36 }对于此类问题，应尽量使用信号量集或者and型信号量以避免死锁。
三、读者写者问题
一个文件，允许多个读者同时读取，只能允许一个写者同时写，写者写的时候不能读。试实现之。
1 using System;
2 using System.Threading;
3
4 namespace ReaderWriter
5 {
6 class ReaderWriter
7 {
8
9 //Warning : Some Error will occur if more than one Reader is newed
10 //Using Semaphore can solve this however here I take mutex as example to show the concept.
11 static int Readcount =0 ;
12 static Mutex WMutex = new Mutex();
13 static Mutex RMutex = new Mutex();
14 static int buffer = 0;
15 static Random rand = new Random();
16
17 static void Main(string[] args)
18 {
19 new Thread(new ThreadStart(Writer)).Start();
20 new Thread(new ThreadStart(Reader)).Start();
21 new Thread(new ThreadStart(Reader)).Start();
22
23
24 Console.Read();
25 }
26
27 private static void Reader()
28 {
29 while (true)
30 {
31 RMutex.WaitOne();
32 if (Readcount == 0)
33 {
34 WMutex.WaitOne();
35 }
36 Readcount++;
37 RMutex.ReleaseMutex();
38
39 Console.WriteLine("I have read this:{0}", buffer.ToString());
40
41 RMutex.WaitOne();
42 Readcount--;
43 if (Readcount == 0)
44 {
45 WMutex.ReleaseMutex();
46 }
47 RMutex.ReleaseMutex();
48 Thread.Sleep(300);
49 }
50 }
51
52 private static void Writer()
53 {
54 while (true)
55 {
56 WMutex.WaitOne();
57 buffer = rand.Next(1, 10);
58 Console.WriteLine("Write {0}",buffer);
59 WMutex.ReleaseMutex();
60 Thread.Sleep(1000);
61 }
62 }
63 }
64 }Rohan
2014-1-7