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经典问题：哲学家就餐问题

经典问题：哲学家就餐问题. 计科 42 1924223 陶益坤 2006.9.29. 临界资源. 问题描述：. 五个哲学家坐在圆桌前，每人一份炒饭； 每个哲学家两侧各有一支筷子； 哲学家处于吃饭和思考两种状态。. 利用记录型信号量解决. #define N 5 void philosopher (int i) { while (true) { 思考； 取 chopstick[i]; 取 chopstick [(i+1) % 5] ； 进食； 放 chopstick [i];

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经典问题：哲学家就餐问题

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  1. 经典问题：哲学家就餐问题 计科42 1924223 陶益坤 2006.9.29

  2. 临界资源 问题描述： 五个哲学家坐在圆桌前，每人一份炒饭； 每个哲学家两侧各有一支筷子； 哲学家处于吃饭和思考两种状态。

  3. 利用记录型信号量解决 #define N 5 void philosopher (int i) { while (true) { 思考； 取chopstick[i]; 取chopstick [(i+1) % 5]； 进食； 放chopstick [i]; 放chopstick [(i+1) % 5]； } }

  4. 为防止死锁发生可采取的措施： • 最多允许4个哲学家同时去拿左边的筷子 • 仅当一个哲学家左右两边的筷子都可用时，才允许他拿筷子（） • 给所有哲学家编号，奇数号的哲学家必须首先拿左边的筷子，偶数号的哲学家则反之

  5. void philosopher (int i) { while (true) { 思考; P(&mutex); state[i] = HUNGRY; test(i); V(&mutex); P(&s[i]); 拿左筷子；拿右筷子； 进食； 放左筷子；放右筷子； P(&mutex) state[ i ] = THINKING； test([i-1] % 5); test([i+1] % 5); V(&mutex); } } state[ i ] = THINKING s[ i ] = 0 #define N 5 #define THINKING 0 #define HUNGRY 1 #define EATING 2 #typedef int semaphore; int state[N]; semaphore mutex=1; semaphore s[N]; void test(int i) { if (state[ i ] == HUNGRY) && (state [ (i-1) % 5] != EATING) && (state [ (i+1) % 5] != EATING) { state[ i ] = EATING; V(&s[ i ]); } }

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