Description: 经典同步问题中的哲学家进餐问题（材料：哲学家5个，筷子5只。要求：每个哲学家必须取得2只筷子时才能进餐，进餐完放下筷子，不能有哲学家饿死。ps：各哲学家每个状态时间有随机性）-Classic synchronous problem :five philosophers have dinner(material:5 philosophers ,five chopsticks.Demand: Each philosophers must have dinner with 2 chopsticks,and lay down the chopsticks after finish eating.No starved philosopher! ) Platform: |
Size: 404521 |
Author:zqm |
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Description: 这是个操作系统中,多线称关于哲学家就餐问题的演示程序!-This is the demo program for the multithread about philosophers having dinner problem in the operating system! Platform: |
Size: 109681 |
Author:张军 |
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Description: 经典同步问题中的哲学家进餐问题（材料：哲学家5个，筷子5只。要求：每个哲学家必须取得2只筷子时才能进餐，进餐完放下筷子，不能有哲学家饿死。ps：各哲学家每个状态时间有随机性）-Classic synchronous problem :five philosophers have dinner(material:5 philosophers ,five chopsticks.Demand: Each philosophers must have dinner with 2 chopsticks,and lay down the chopsticks after finish eating.No starved philosopher! ) Platform: |
Size: 404480 |
Author:zqm |
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Description: 这是个操作系统中,多线称关于哲学家就餐问题的演示程序!-This is the demo program for the multithread about philosophers having dinner problem in the operating system! Platform: |
Size: 123904 |
Author:张军 |
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Description: 进程互斥进程同步
哲学家有N个，也定全体到达后开始讨论：在讨论的间隙哲学家进餐，每人进餐时都需使用刀、叉各一把，所有哲学家刀和叉都拿到后才能进餐。哲学家的人数、餐桌上的布置自行设定，实现刀和叉的互斥使用算法的程序实现。-The philosopher N, also all arrived to begin to discuss: in the discussion of the gap the dining philosophers, each table needs to use knife, fork of a knife and fork, all the philosophers have only after dinner. The number of philosophers, table layout set, to achieve a knife and fork of the exclusive use of the program algorithm. Platform: |
Size: 2048 |
Author:赛德华 |
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Description: 哲学家就餐问题的仿真，仿真结果给出每个哲学家等待的平均时间。此外还有多种策略的对比，以评判策略的优劣。-Simulation of the philosophers s repast problem, the simulation gives the average time that philosopher has been waited for dinner. In addition there are a variety of strategies, then it can judge the merits of different strategies. Platform: |
Size: 479232 |
Author:尹龑燊 |
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Description: 利用JAVA线程，解决哲学家就餐问题。当某一哲学家线程执行取得筷子方法时， 程序会根据该线程的名称来确定该线程需要使用哪两支筷子，并且分辨出哪支筷子编号是奇数，按照先奇后偶的顺序来试图取得这两支筷子。 如果这两支筷子都未被使用(即对应的数组元素值为 false)，该哲学家线程即可先后取得这两支筷子进餐，否则会在竞争某支筷子失 败后执行 wait()操作进入 Chopsticks 类实例的等待区， 直到其他的哲学家线程进餐完毕放下筷子时用 notifyAll()将其唤醒。当某一哲学家线程放下筷子时， 程序会将放下的筷子对应的数组元素值置为 false，并用 notifyAll()唤醒在等待区里的其他线程。(Use the JAVA thread to solve the problem of the philosopher's eating. When a philosopher thread performs the way of chopsticks, the program will determine which two chopsticks he needs to use according to the name of the thread, and identify which chopsticks are odd numbers, and try to get the two chopsticks according to the order of odd and even later. If the two chopsticks are not being used (i.e. the corresponding array element value of false), the philosopher has made this thread can be two chopsticks, otherwise it will execute wait in the competition after the failure of chopsticks () into the operation of an instance of the Chopsticks class to wait until the other thread, put down the chopsticks when philosophers after dinner (notifyAll) in its wake. When a philosopher thread puts down the chopsticks, the program sets the value of the array elements corresponding to the dropped chopsticks into false, and wakes up the other threads in the waiting area with notifyAll ().) Platform: |
Size: 67584 |
Author:打打打、
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