用加锁的方法解决两个进程累加数组的互斥问题-locked with two ways to solve the array of cumulative process Mutual Exclusion

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

进程管理中的生产者消费者问题 只要是是描述进程管理中生产者与消费者在共享文件时是否同步与互斥问题-management process, the producer consumer issues as long as it is described process management producers and consumers when sharing files synchronized with the Mutual Exclusion

哲学家进餐问题，描述计算机系统的同步与互斥问题。-dining philosopher, describes the computer system synchronization and mutual exclusion problems.