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Philosopher(new)
- 这是一道经典的线程同步问题——哲学家吃饭问题。这里结合了win32使得这个问题可视化了-This is a classic thread synchronization -- philosopher food problem. Win32 here combination of this makes the issue of visualization
chip(单片机源程序)
- 好东西 关于单片机的一些重要的代码本程序是操作系统中比较典型的线程同步算法中的哲学家进餐问题,为防止死锁,采取了两个条件(筷子空闲)同时满足时再允许进餐的办法来解决。期望与大家一起学习交流!-good things on SCM some important source of this procedure is more typical operating system threads simultaneously count the law on the issue of the Dinin
PhilosopherDining
- windowsAPI编程实现哲学家用餐问题。 五个哲学家五个线程。监测死锁并解决-windowsAPI Programming dining philosophers problem. 5 philosopher five threads. Monitoring and addressing Deadlock
CrazyPhilosopher
- Crazy Philosopher 演示关于操作系统多线程的经典的“哲学家吃饭问题”。-Crazy Philosopher demonstration multithreaded operating system on the classic "philosopher eat rice problem. "
threadphilosopher
- 利用线程方案实现实现哲学家问题.设置五个信号量代表五把叉子,初使值均为1,表示5把叉子均可以取,设置五个线程代表5个哲学家,其值分别为0~4,规定奇数号的哲学家先拿起他左边的叉子,然后再去拿他右边的叉子 而偶数号的哲学家则相反.按此规定,将是1,2号哲学家竞争1号叉子,3,4号哲学家竞争3号叉子.即五个哲学家都竞争奇数号叉子,获得后,再去竞争偶数号叉子,最后总会有一个哲学家能获得两支叉子而进餐。而申请不到的哲学家进入阻塞等待队列,根FIFO原则,则先申请的哲学家会较先可以吃饭,因此不会出现饿死的
Philosophers
- 哲学家进餐问题,进程是独立参与分配资源的最小单位,在有线程的OS中,线程是运行的最小单位, 课堂所述进程之间的同步与互斥,实际上是属于不同进程的线程间的同步与互斥,当 然,属于同一进程的不同线程一样存在同步与互斥,其控制同步与互斥的原理跟进程 是一样的-dining philosophers problem, the process is independent participation in the allocation of resources as the smallest
lunch
- 哲学家进餐问题 在多线程中如何避免死锁。 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 要么等待,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-dining philosophers problem in multithreaded how to avoid deadlock. Problem descr iption : five philosophers sittin
process_tongbu
- 用win32函数,模拟了哲学家进餐问题,对于初学多线程编程的理解线程之间的同步和死锁问题很有启发性。-using win32 function to simulate the Dining Philosophers, For novice multithreaded programming understanding thread synchronization between the deadlock and very instructive.
jincan
- 哲学家进餐问题 在多线程中如何避免死锁。 简单源码,有参考价自己 -dining philosophers problem in multithreaded how to avoid deadlock. Simple source, a reference price of its own
哲学家进餐问题
- 程序是操作系统中比较典型的线程同步算法中的哲学家进餐问题,为防止死锁,采取了两个条件(筷子空闲)同时满足时再允许进餐的办法来解决。
哲学家进餐问题VC版
- 本程序是操作系统中比较典型的线程同步算法中的哲学家进餐问题,为防止死锁,采取了两个条件(筷子空闲)同时满足时再允许进餐的办法来解决。期望与大家一起学习交流!-the procedure is fairly typical operating system threads synchronization algorithm of the Dining Philosophers, in order to prevent deadlock and take the two conditions (ch
dining_philosopher.tar
- 哲学家进餐经典算法的pthread多线程实现代码-Classic dining philosophers pthread algorithm to achieve multi-threaded code
java_threaddemo
- 哲学家就餐是一个经典的Java多线程编程的实例,这是图形版。涉及到线程同步与互斥,临界区访问问题以及避免死锁的方法。 -Dining philosophers is a classic multi-threaded Java programming examples, this is the graphics version. Related to thread synchronization and mutual exclusion, critical area to visit, as we
Dining_philosophers
- 用VC实现的哲学家就餐问题,其中用到了信号量的宏,并多线程的方法,模拟哲学家就餐问题。-Implementation using VC dining philosophers problem, use one of the semaphore macros, and multi-threaded approach, Analog dining philosophers problem.
java_threaddemo
- 哲学家就餐是一个经典的Java多线程编程的实例,这是图形版。涉及到线程同步与互斥,临界区访问问题以及避免死锁的方法。 -Dining Philosophers is a classic example of Java multi-threaded programming, it is graphic version. Related to thread synchronization and mutual exclusion, critical access issues and ways to
operation
- c++实现的进程线程问题,包括父子进程,哲学家进餐问题等等.-c++ implementation of the process of threading issues, including his son process, the philosopher eating problems, and so.
chopsticktest
- 这是一个用多线程编写的哲学家进餐问题,里面用到了多线程中的一些知识。-This is a multi-threaded, prepared with the dining philosophers problem, which uses a multi-threaded some knowledge.
philpsopher
- 哲学家就餐问题的解决方法 运用java多线程模拟所有解决哲学家就餐问题的方法-Solution to the dining philosophers problem using multi-threaded simulation of java all the dining philosophers problem solving
thinking
- 根据哲学家进程间的相互制约关系,设置合理的信号量及信号量初值。 创建5个线程,分别模拟5个哲学家进程。 在哲学家线程之间通过对信号量的P,V操作,实现线程之间的同步关系。(According to the mutual restriction between the philosophers' process, the reasonable signal quantity and the initial value of the signal quantity are set up. Cr
多线程调度——哲学家就餐
- 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