当前位置: 首页 > 图文教程 > 网络编程 > ASP.NET > 算法讨论:哲学家就餐问题

ASP.NET
调用另外一个同名的重载函数漏掉括号出现的严重问题
DataGrid模板列应用:在DataGrid中用CheckBox控制TextBox的Enabled属性
使用存储过程的一个小例子
.net中xmlhttp下载文件的方法参考
简单的c#文本文件读写
我的C#学习过程 第一天 安装
在数据库中开始一个事务。
用ASP.NETt实现简单的文字水印
C/S系统中三层结构(Com/Com+)的测试成功实现
VB 二进制块读写类模块应用实例,包括一个文件拷贝和一个文件二进制比较的例子。
扫雷程序“布雷”代码(vb)
关于网络连接状态的编程
c#中ref和out参数使用时需要注意的问题
use Assembly to call a method
在VBA中调用AUTOCAD打印文件
在VS.NET的
昨天折腾了一晚上,哪位高手帮我看看!急!
Net是未来的趋势吗?
如何修改自定义Webpart 的标题(downmoon)
[收藏]ASP.Net生成静态HTML页 选择自 coofucoo 的 Blog

ASP.NET 中的 算法讨论:哲学家就餐问题


出处:互联网   整理: 软晨网(RuanChen.com)   发布: 2009-11-03   浏览: 76 ::
收藏到网摘: n/a

In 1965, Dijkstra posed and solved a synchronization problem he called thedining philosophers problem. ........ The problem can be stated quite simplyas follows. Five philosophers are seated around a circular table. Each philosopher has a plate of spaghetti. The spaghetti is so slippery that a philosopher needs two forks to eat it. Between each pair of plates is one fork. The life of a philosopher consists of alternate periods of eating and thinking. When a philosopher gets hungry, she tries to acquire her left and rightfork, one at a time, in either order. If successful in acquiring two forks,she eats for a while, then puts down the forks and continues to think. Thekey question is: Can you write a program for each philosopher that does what it is supposed to do and never gets stuck? --from written by Andrew S. Tanenbaum typed by foolball :-PProgramme provided by ya: : 法一: 用公共文件,按照严格轮流执行: : #include : : #include : : #include : : #include : : #define N 5: : int i,j,t,status;: : FILE * f;: : char *state[N];: : main(): : {: : f=fopen("/share","w+");: : putc(j,f);: : if (fork()): : { if (fork()): : { if (fork()): : { if (fork()): : { if (fork()): : {waitpid(-1,*status,0);: : fclose(f);}: : else : : philosophy(4);}: : else: : philosophy(3);}: : else: : philosophy(2);}: : else: : philosophy(1);}: : else: : philosophy(0);: : }: : void philosophy(int i): : {: : state[i ="thinking";: : printf("%d%s\n",i,"is thinking");: : for(t=0;t<=rand()+10000;t++);: : state[i = : : printf("%d%s\n",i,"is hungry");: : for(t=0;t<=rand()+10000;t++);: : for(;i!=j;): : {fseek(f,0l,0);: : j=getc(f);} : : state[i ="eating";: : printf("%d%s\n",i,"is eating");: : j=(j+1)% N: : fseek(f,0l,0);: : putc(j,f);: : }: : 法二:通过文件加锁实现: : #include : : #include : : #define N 4: : FILE *f;: : int i, status;: : char *state[N];: : void philosofy(int i);: : void main(): : {: : if ((f=fopen("turn", "w+"))==NULL): : {: : printf("Cann't open this file"); : : exit(0);: : }: : if (fork()): : {: : if(fork()): : {: : if(fork()): : {: : if(fork()): : {: : if(fork()): : {: : waitpid(-1, &status, 0);: : fclose(f);: : }: : else: : philosofy(4);: : }: : else: : philosofy(3);: : }: : else : : philosofy(2);: : }: : else: : philosofy(1);: : }: : else: : philosofy(0);: : }//end of main: : void philosophy(int i): : { int t;: : state[i ="thinking";: : printf("%d%s\n",i,"is thinking");: : for(t=0;t<=rand()+10000;t++);: : state[i ="hungry";: : printf("%d%s\n",i,"is hungry");: : for(t=0;t<=rand()+10000;t++);: : while ((f=fopen("turn.lock","r"))!=NULL);: : link ("turn","turn.lock");: : state[i ="eating"; : : printf("%d%s\n",i," is eating");: : for(t=1; t<=10000+rand();t++) ;: : unlink ("turn");: : }