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philosopher
- 用图形界面模拟如下场景：一个圆桌上有一个大碗面，5个盘子，每位哲学家一个，还有5把叉子。每个想吃饭的哲学家将做到桌子旁边分配给他的位置上，使用盘子两侧的叉子，取面和吃面。问题是：设计一个算法以允许哲学家吃饭。算法必须保证互斥（没有两位哲学家同时使用同一把叉子），同时还要避免死锁和饥饿。-Use graphical interface simulation scenario is as follows: a round-table, th
philosopher
- 五个哲学家的就餐问题，程序中有左撇子存在，不存在饿死的情况！-Five dining philosophers problem, procedures exist in the left-handed, there is no case of starvation!
eat
- 哲学家进餐问题，经典的同步问题，是初学者很好的学习材料-Dining philosophers problem, the classical synchronization problem is a very good learning materials for beginners
DINING
- 演示哲学家就餐 进程的同步与互斥-Demo dining philosophers process synchronization and mutual exclusion
eating
- 模拟哲学家进餐的同步问题，实现一个既没有两邻座同时吃饭，又没有人饿死（永远拿不到一双筷子）的算法-Simulation of synchronous dining philosophers problem, the realization of a neighbor at the same time do not have two meals, and no one starved to death (never get a pair
ucos-dining-philosopher
- ucos-dining-philosopher.rar ucos下的哲学家就餐问题（dining philosopher problems)，采用信号量机制解决，有截图和感悟。dos界面文字演示（另有图形界面）。-ucos-dining-philosopher.rarucos under the dining philosophers problem (dining philosopher problems), using sema
philosophers
- typical problem of the philosophers
Philosophers
- 哲学家就餐问题，他们每个人的左边都有一支筷子，解决死锁问题-Dining philosophers problem, the left side of each of them has a chopsticks, solve the deadlock problem
Philosophers
- Dining-Philosophers Problem
exp4
- 哲学家就餐问题C语言模拟，pthread库实现-The Dining Philosophers Problem C language simulation, pthread library implementation
Dining
- 程序源代码： /* Item The Dining Philosophers Problem 具体要求 1) 5个哲学家，先吃饭，后思考。 2) 每个哲学家都需要吃9顿饭，思考9次，然后结束。 3）吃饭时间为3~12秒的随机时间 4）思考时间为7~19秒的随机时间 5）对取叉子和放叉子等临界资源的操作前后需要输出相关状态信息，包括哲学家的各种状态信息。要有足
phenosophy
- 本程序是在Linux下面 实现了哲学家就餐问题。其中演示了会发生死锁和不会发生死锁的情况。-This program is implemented in the Linux following the Dining Philosophers Problem. Which demonstrates the deadlock and does not happen will happen deadlock situation.
Philosophers
- 这个程序可以演示关于哲学家就餐的问题，对于理解线程的相互等待有很好的作用-This procedure can be demonstrated on the Dining Philosophers problem, each thread to wait for understanding the role of good
philosophers
- linux下哲学家用餐的经典问题，内含makeifile和主函数源代码-Under the classic dining philosophers linux problem and the main function contains the source code makeifile
dining
- 哲学家进餐代码 附加可执行文件 哲学家进餐代码 附加可执行文件-Dining philosophers code attached executable file attached executable code dining philosophers
Dining-philosophers-problem
- 哲学家就餐问题 实现5个哲学家5支筷子的就餐问题涉及到AND型信号量，线程的阻塞唤醒切换功能-Dining philosophers problem to achieve five dining philosophers five chopsticks issues related to the AND-type semaphore, blocking thread switching wake
Dinning-philosophers
- 在linux下“哲学家就餐”问题的一种解法-a solution of "dining philosophers" problem under Linux
[java]philosophers-eating
- 本java设计模拟经典的哲学家进餐的同步问题，实现一个既没有两邻座同时吃饭，又没有人饿死（永远拿不到一双筷子）的算法。内附运行说明。-The Java design simulation of the classic dining philosophers synchronization issues, to achieve a two sitting next to him to eat, and no one starved to
philosophers
- 通过实现哲学家进餐问题的同步深入了解和掌握进程同步和互斥的原理-Through the dining philosophers problem of synchronization-depth understanding and mastery of process synchronization and mutual exclusion principle
philosophers
- 哲学家进餐例子。 哲学家问题的大意是：5个哲学家围坐在一张圆桌周围，每个哲学家面 前有一碟通心面,要使用筷子才能进餐。相邻两个碟子之间有一根筷子。哲 学家包含三种状态：吃饭、思考和挨饿。当一个哲学家觉得饿时，他才试 图分两次去取他左边和右边的筷子就餐，吃完以后放下筷子继续思考。-The dining philosophers example. Philosophers problem to the effect that: