创建两个生产者进程和两个消费者进程，生产者进程a需要生成10000个整数,每次都将自己的进程号（用getpid()函数获得）和生成的整数放入共享内存中（共享内存大小为64Byte）。生产者b每次从26个英文字母中选一个，并将自己的进程号和选中的字母放入共享内存中，直到26个字母全部都选中。消费者进程c负责从共享内存中读取数据生产者进程a的数据并且将这些数据写入文件a.out。消费者进程d从共享内存读取进程b的数据后写入b.out中。
标签： 进程
上传时间： 2014-01-24
上传用户：王者A
本课题设计的双机通信系统，应能完成2台80X86PC机的串行通信。将数据从A机发送到B机，或从B机发送到A机。
标签： 双机通信
上传时间： 2014-07-13
上传用户：stampede
This article describes how to consume Web services in Java 1.5.0 using the new JAX-WS 2.0 API (JSR 228). Developers around the world, including me, have always complained about the hard ways to work in Java to consume even a Web service as simple as adding two numbers. However, with JAX-WS 2.0 API now available in core Java in JDK 1.5.0, life is simple like never before.
标签： describes services article consume
上传时间： 2013-11-27
上传用户：kelimu
3G Mobile Open Wide Door For E-commerce The 3G mobile Internet business in 3G era will obtain rapid development, but it still cannot become the ma in 3G era. In the age of 3G speech business, but is still a subject of value-added business will have great development, the 3G mobile Internet business.
标签： E-commerce 3G Internet business
上传时间： 2017-03-23
上传用户：zjf3110
In computer vision, sets of data acquired by sampling the same scene or object at different times, or from different perspectives, will be in different coordinate systems. Image registration is the process of transforming the different sets of data into one coordinate system. Registration is necessary in order to be able to compare or integrate the data obtained from different measurements. Image registration is the process of transforming the different sets of data into one coordinate system. To be precise it involves finding transformations that relate spatial information conveyed in one image to that in another or in physical space. Image registration is performed on a series of at least two images, where one of these images is the reference image to which all the others will be registered. The other images are referred to as target images.
标签： different computer acquired sampling
上传时间： 2013-12-28
上传用户：来茴
one of video tool Skype only in windows mobile5.0 more than running, but my task is to let it run Skype in WINCE only in windows mobile5.0 more than running, but my task is to let it run in WINCE . Leave no stone unturned when I can not get windows mobile, they thought would be needed for skype all library files to my WINCE, so perhaps will be able to run skype. 1. Used depend Show skype depends on which files. In fact, not many documents, I can remember only under the following: core.dllcommctr.dllaygeshell.dllhtmlview.dll In addition to htmlview.dll, the other documents in WINCE have (if your system does not, please IMAGE Canada on the corresponding module). Therefore, I find on the Internet htmlview.dll, the results really find
标签： running windows mobile Skype
上传时间： 2014-01-17
上传用户：miaochun888
编写一个java应用程序。用户从键盘输入一个1-9999之间的数，程序将判断这个数是几位数，并判断这个数是否回文数。回文数是指将数含有的数字逆序排列后得到的数和原数相同，例如12121,4224,6778776等都是回文数。 1）程序具有判断用户的输入是否为合法整数的功能。对非法输入（例如含有字母）要进行处理。 2）要判断输入数的位数，并输出相关信息。 3）要判断是否回文数。 二、二战期间，英国情报人员获取德军的一机密电报，电报的内容为： bzdz izu sxgzd vs lh ,vpzg woflsh vs vwrh vhlsddlmp glm wrw gzy vsg .gflyz gstfzu bvsg gzsd hdmlp vml lm ,hghzvy wmz hwiry mvvdgvy izd z hzd vivsg ,ltz tmlo tmlO 情报人员已经知道，这段电报的加密方式为： 1. 首先将字符串的顺序颠倒。 2. 字母互换的规律为：A->Z, B-Y, C-X...X->C, Y->B, Z-A a->z, b->y, c-x...x->c, y->b, z->a. 3. 非字母字符保持不变。 请编程帮助情报人员破译这份机密电报。给出注释良好的源程序和程序运行后的结果。
上传时间： 2017-06-02
上传用户：dengzb84
ZEUS is a family of Eulerian (grid based) Magneto-Hydrodynamic codes (MHD) for use in astrophysics, described in a series of papers by Stone and Norman (1, 2, 3). It may be used in Cartesian (XYZ), cylindrical (ZRP), and spherical (RTP) coordinates.
标签： Magneto-Hydrodynamic astrophysics Eulerian family
上传时间： 2013-12-19
上传用户：maizezhen
两数比较大小，交换以保持a>b 如果a已经大于b则不交换 交换前后状态都显示
标签： 比较
上传时间： 2014-12-03
上传用户：luke5347
实验源代码 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("请输入矩阵第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可传递闭包关系矩阵是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元关系的可传递闭包\n"); void warshall(int,int); int k , n; printf("请输入矩阵的行数 i: "); scanf("%d",&k); 四川大学实验报告 printf("请输入矩阵的列数 j: "); scanf("%d",&n); warshall(k,n); }
上传时间： 2016-06-27
上传用户：梁雪文以
