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实验四 模拟“五个哲学家”问题
- 里面有代码跟实验报告
Stm32_ucos 在STM32上编写多任务实时操作系统
- 在STM32上编写多任务实时操作系统，同时运行6个任务，通过发送和接收信号量机制，解决哲学家就餐问题，通过板上按键代表哲学家的就餐请求，板上LED灯表示就餐成功或者未果，同时在晶体数码管上显示就餐申请时间。-STM32 to write multi-task in real-time operating system, run six tasks by sending and receiving semaphore mechanism to solve the dining philosophe
zhexuejia
- C多线实现哲学家就餐问题-C multi-line to achieve the dining philosophers problem
success3
- 哲学家问题，C++编的多线程模拟哲学家吃饭问题，运行前需要设置一下project-Philosophers problem, C++ code for multi-threaded simulation of philosophers problem of feeding the project need to set before running
philosopher
- 采用c语言实现哲学家进餐问题，实现进程机制的管理！-Using c language implementation philosopher dine and realize process of management mechanism
DiningPhilosophers
- Linux下多线程实现哲学家进餐算法。使用了线程间协调通信，有优先级，防止了死锁。-Dining philosophers algorithm Multithread was utilized. Compiled in Linux GCC
operation-2
- 哲学家就餐问题--操作系统理解pv原语 解决进程间同步问题-operation problem of Dining Philosophers
phy
- 多线程解决哲学家就餐问题，利用操作系统中互斥量和信号量-Multi-threading to solve the dining philosophers problem, using the operating system mutex and semaphore
TC-ucos-philos-comment-in-detail
- TC-ucos-philos-详细注释.rar 用TC编译ucos并在图像化界面下演示哲学家就餐问题，采用信号量机制，程序有BUG，不能顺利运行，但是我写了很多注释，有兴趣的朋友可以继续做下去，我的邮箱xuxi009@163.com可与我联系。注意图形界面驱动的路径要和源文件中的相同否则无法显示图形界面，相信用过TC的图形界面的明白，呵呵。-TC-ucos-philos-detailed notes. Rar uCOS compiled by TC and the image of i
ARM-UCOS-philos
- 在S3C44b0上移植ucos并实现哲学家就餐问题的演示，信息从uart0口输出，这是我学习上课时做的一个实验，见笑见笑了-S3C44B0 uCOS transplantation in the dining philosophers problem and the realization of the demonstration, information from the UART0 port output, this is my learning in class to do an expe
eating
- 哲学家就餐问题 哲学家就餐问题 哲学家就餐问题-Dining philosophers problem Dining philosophers problem Dining philosophers problem
zhexuejia
- 计算机操作系统，哲学家就餐问题，C源代码。-Computer operating system, philosophers eat, C source code.
zhwxuejia
- 一群哲学家围坐在一个圆桌，手上持有密码m，并从1开始编了号取初值m，哲学家从1开始报数, 报到m的哲学家停止吃饭，退出圆桌，求哲学家退出的顺序。要求:n和初值m由完家输入.手上的密码随机产生.最后要打印出编号对应的密码，输出哲学家离开的相后顺序-A group of philosophers sitting in a round-table, holding hands password m, and from the No. 1 started to take the initial valu
001
- 一群哲学家围坐在一个圆桌，手上持有密码m，并从1开始编了号取初值m，哲学家从1开始报数, 报到m的哲学家停止吃饭，退出圆桌，求哲学家退出的顺序。要求:n和初值m由完家输入.手上的密码随机产生.最后要打印出编号对应的密码，输出哲学家离开的相后顺序-A group of philosophers sitting in a round-table, holding hands password m, and from the No. 1 started to take the initial val
200811182357586855
- 一群哲学家围坐在一个圆桌，手上持有密码m，并从1开始编了号取初值m，哲学家从1开始报数, 报到m的哲学家停止吃饭，退出圆桌，求哲学家退出的顺序。要求:n和初值m由完家输入.手上的密码随机产生.最后要打印出编号对应的密码，输出哲学家离开的相后顺序-A group of philosophers sitting in a round-table, holding hands password m, and from the No. 1 started to take the initial val
002
- 一群哲学家围坐在一个圆桌，手上持有密码m，并从1开始编了号取初值m，哲学家从1开始报数, 报到m的哲学家停止吃饭，退出圆桌，求哲学家退出的顺序。要求:n和初值m由完家输入.手上的密码随机产生.最后要打印出编号对应的密码，输出哲学家离开的相后顺序-A group of philosophers sitting in a round-table, holding hands password m, and from the No. 1 started to take the initial valu
philosopher
- 哲学家就餐问题的实现，进程死锁问题，PV原语-Philosophers eat the realization of the process of deadlock, PV primitives
ex3
- 哲学家就餐问题的实现！是使用信号量实现的，有利于ucos的学习-Implementation of the dining philosophers problem! Is to use semaphore implementation, and is conducive to learning uCOS
PE0213
- 操作系统中的哲学家就餐问题模拟系统。可视化界面程序。c#实现。运行需要.NET Framework 2.0。另有命令行程序。-Operating system in the dining philosophers problem simulation system. Visualization interface procedures. c# implementation. Required to run. NET Framework 2.0. Another command-line prog
Dining_Philosophers
- 英国帝国理工学院电脑系多线程课程教学材料之八 ---- 哲学家晚餐范例-Five philosophers sit around a circular table. Each philosopher spends his life alternatively thinking and eating. In the centre of the table is a large plate of spaghetti. A philosopher needs two forks to eat a he