增加哲学家就餐问题+代码

Operating Systems.md

@@ -100,47 +100,91 @@ wait操作：执行wait操作的进程进入条件变量链末尾，唤醒紧急
 <summary>生产者-消费者问题</summary>
 
 > 问题描述：使用一个缓冲区来存放数据，只有缓冲区没有满，生产者才可以写入数据；只有缓冲区不为空，消费者才可以读出数据
->
-> ```c
-> // 伪代码描述 
-> // 定义信号量 full记录缓冲区物品数量 empty代表缓冲区空位数量 mutex为互斥量
-> semaphore full = 0, empty = n, mutex = 1;
-> 
-> // 生产者进程
-> void producer(){
-> 	do{
->    	  P(empty);
-> 	  P(mutex);
-> 
->      // 生产者进行生产
->    	
->    	  V(mutex);
->    	  V(full);
->  	} while(1);
-> }
-> 
-> void consumer(){
-> 	do{
-> 	  P(full);
-> 	  P(mutex);
-> 
->     	// 消费者进行消费
-> 
-> 	  V(mutex);
-> 	  V(empty);
->  	} while(1);
-> }
-> 
-> ```
->
-> 
->
-> </details>
+
+代码实现：
+```c
+// 伪代码描述 
+// 定义信号量 full记录缓冲区物品数量 empty代表缓冲区空位数量 mutex为互斥量
+semaphore full = 0, empty = n, mutex = 1;
+
+// 生产者进程
+void producer(){
+	do{
+   	  P(empty);
+	  P(mutex);
+
+     // 生产者进行生产
+
+   	  V(mutex);
+   	  V(full);
+ 	} while(1);
+}
+
+void consumer(){
+	do{
+	  P(full);
+	  P(mutex);
+
+    	// 消费者进行消费
+
+	  V(mutex);
+	  V(empty);
+ 	} while(1);
+}
+
+```
+</details>
 
 <details>
 <summary>哲学家就餐问题</summary>
 
-
+> 问题描述：有五位哲学家围绕着餐桌坐，每一位哲学家要么思考，要么吃饭。为了吃饭，哲学家必须拿起两双筷子（分别放于左右两端）不幸的是，筷子的数量和哲学家相等，所以每只筷子必须由两位哲学家共享。
+
+代码实现：
+```c
+#define N 5  // number of philosopher
+#define LEFT (i + N - 1)%N // number of i's left neighbors
+#define RIGHT (i + 1)%N // number of i's right neighbors
+#define THINKING 0
+#define HUNGRY 1
+#define EATING 2
+typedef int semaphore;
+int state[N]; // array to keep track of everyone's state
+semaphore mutex = 1; // mutual exclusion of critical region
+semaphore s[N]; 
+
+void philosopher(int i) {
+	while (TRUE) {
+		think();
+		take_forks(i);
+		eat();
+		put_forks(i);
+	}
+}
+
+void take_forks(int i) {
+	down(&mutex); // enter critical region
+	state[i] = HUNGRY; // record that i is hungry
+	test_forks(i); // try to acquire two forks
+	up(&mutex); // exit critical region
+	down(&s[i]); // block if forks are not acquired
+}
+
+void put_forks(int i) {
+	down(&mutex); // enter critical region
+	state[i] = THINKING; // record that has finished eating
+	test_forks(LEFT); // see if left neighbor can now eat
+	test_forks(RIGHT); // see if right neighbor can now eat
+	up(&mutex); // exit critical region
+}
+
+void test_forks(int i) {
+	if (state[i] == HUNGRY && state[LEFT] != EATING && state[RIGHT] != EATING) {
+		state[i] = EATING;
+		up(&s[i]);
+	}
+}
+```
 </details>
 
 <details>