信号量 进程 (m个生产者,n个消费者,容量为r的缓冲区)
来源:互联网 发布:html5 php区别 编辑:程序博客网 时间:2024/06/15 23:00
转载,原文地址http://www.cnblogs.com/phinecos/archive/2006/08/25/486552.html
1.整型信号量是一个整数变量,除初始化外,对其只能执行两个操作,即wait(s)和signal(s),也叫p(s)和v(s)操作,均是原语操作,用来实现进程的同步,互斥.
2.记录型信号量
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结论:
(1) 若信号量s为正值,则等于在封锁进程之前对信号梁可以施行的p操作数,也就是s所代表的实际可以使用的物理资源数.
(2) 若信号量s为负值,则其绝对值等于排列在该信号量s队列之中等待的进程个数,也就等于对信号量s实施p操作而被封锁起来并进入信号量s队列的进程数.
(3)p操作一般代表请求一个资源,v操作一般代表着释放一个资源,在一定条件下,p操作代表挂起进程操作,v操作代表唤醒被挂起的进程.
3.经典同步问题
1)生产者/消费者问题
a) 问题一:有一个生产者和一个消费者,共享一个缓冲区.两者要互斥访问缓冲区.
解:定义两个信号量:empty表示缓冲是否为空,初值为1,即初始时可以存入一件物品.full表示缓冲区中是否有物品,初值为0,即初始时缓冲区没有物品.
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注:
(1)进程互斥只需要一个信号量,而同步可能要两个信号量.
(2)p,v操作仍然要成对,但在进程进入临界区前后调用的是针对不同信号量的wait,signal操作,而进程互斥时是针对相同的信号量.
(3)至少有一个信号量的初值>=1,否则所有进程无法执行,一般是指管理是否允许访问共享资源的那个信号量.如这里的empty设为1,如果缓冲区容量为n,则可以设为n;
b)问题二:m个生产者,n个消费者,容量为r的缓冲区,(m,n,r都大于1),不要生产者和消费者互斥存取物品.
解:1)生产者和消费者之间要同步,类似问题一,用两个信号量empty,full;2)m个生产者之间要互斥,n个消费者之间也要互斥,但生产者和消费者不用互斥存取物品,因此设两个计数器in,out和相应的互斥信号量mutex1,mutex2
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c)问题三:m个生产者,n个消费者,容量为r的缓冲区,(m,n,r都大于1),要求生产者和消费者互斥存取物品.
解:1)生产者和消费者之间要同步,类似问题一,用两个信号量empty,full;2)m个生产者之间要互斥,n个消费者之间也要互斥,同时要求生产者和消费者互斥存取物品,因此设两个计数器in,out和一个互斥信号量mutex.
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注:应该先执行对资源信号量的p,v操作,再执行互斥信号量的p,v操作,否则会引起死锁.
2)读者/写者问题
要求:(1)允许多个读者同时从数据区读数据;
(2)当读者在读数据时,不允许写者写数据
(3)任何时候只允许一个写者向数据区写数据;
(4)若有写者在写数据,则不允许读者读数据.
问题分为读者优先和写者优先两类。若有读者在读数据时,写者到来时就会被阻塞,直到所有读者读完数据才被唤醒.这里读者优先于写者,称为第一类读写问题.若读者在读数据时,有写者到来,则后续的读者将被阻塞,直到写者离开才被唤醒,这类问题中写者处于优势地位,叫第二类读写问题.
a)问题一:用记录型信号量解第一类读写问题
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注:
(1)WRmutex用于读者和写者,写者和写者之间互斥访问数据区.
(2)当没有读者访问时,允许一个读者进入,第一个读者进入时要用WRmutex与写者互斥.
(3) readcount记录正在读数据区的读者数量,由于能被多个读者进程共享,也是临界资源,因此用Rmutex来对其实施互斥访问.
(4)允许多个读者同时读数据,当至少有一个读者在读时,其他读者不用等待就可以读数据。
(5)一旦有一个读者开始读数据,其后,只要至少还有一个读者在读数据,读者就一直保持对数据区的控制权,这会造成写者的"饥饿";
b)问题二:用AND信号量解第一类读写问题
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c)第二类读写问题
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3)哲学家进餐问题
a)使用记录型信号量
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b)使用AND信号量
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4.真题精选
[复旦2000]
问答: 在计算机科学文献中的几种互斥算法中,所谓的Lamport面包店算法可以有效地用于多个相互竞争的控制线程,该算法中线程之间的通信只能在共享内存中进行(即,不需要诸如信号量、原子性的set-and-test之类的专门机制)。该算法的基本思想源于面包店,因为面包店需要先取号然后等候叫号。下面给出了该算法的框架,该算法可以使各线程进出临界区而不产生冲突。
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定义(a,b)<(c,d)为(a<c) or (a=c and b<d);
Enter[i]和Number[i]可以被进程i读写,但是只能被进程j(j<>i)读.
请问:(1)算法如何解决互斥访问临界区问题
(2) 算法如何解决死锁问题
分析:每个进入面包店的顾客会得到一个号码,服务员为拥有最小号码的顾客服务.此算法需要进程pi经过两个步骤进入临界区.第一步,它需要选择一个号码,因此,它读取其他进程的号码,并选择一个比它们的最大值大1的数字作为自己的号码(称为"面包店入口).第二步,进程pi用如下方法检查是否可以进入临界区.对任意其他进程pj,pi首先检查pj是否已经在面包店入口,如果在,那么pi等待pj离开面包店入口,然后pi等待Number[j]变为0或者Number[j],j) < (Number[i],i).当pi对所有其他进程成功验证上述条件后,就可以进入临界区.
解:
首先证明两个结论:
(1)若进程pi已经在临界区,而且若干其他进程pk已经选择了他们的号码,那么(Number[i],i) < (Number[k],k).
证明:若pi已经在临界区,则一定经历了k轮for循环,就是说在上述几轮循环中,Number[k]=0或者(Number[i],i) < (Number[k],k).首先假设进程pi读出Number[k]为0,也就是说pk还没有选择好一个号码,这里有两种情形:一是pk不在面包店入口,二是已经进入但还没有退出.若pk不在面包店入口,则它将可以读到最新的Number[i]的值,从而确保Number[k]>Number[i].若它已经在入口处,则此次进入一定是在pi检查了Enter[k]之后,因为pi在检查条件(Number[k]==0) && ((Number[i],i) < (Number[k],k))) 之前需要等待pk完成选择,也就是说pk将读到最新的Number[i]的值,因此确保Number[k]>Number[i].如果在k轮循环中,(Number[i],i) < (Number[k],k).那么这个结论将继续保持,因为Number[i]的值不会改变,而Number[k]只会增加.
(2)若进程pi在临界区中,则Number[i]>0
证明:因为此数值至少为0,当一个进程进入临界区之前一定执行了一个加1的操作.
由上述结论,若有两个进程pi和pk同时进入了临界区,由(2)知它们的号码都大于0,由(1),则有Number[k]>Number[i],反之也一样,所以矛盾.
对于死锁问题,由上述结论知,任一时刻一定存在一个进程pi,它的(Number[i],i)值最小,因此它总能够执行完成并归还资源,所以不会产生死锁.
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