词法分析 1 试验目的 设计,编制并调试一个此法分析程序,加深对此法分原理的理解. 2 试验要求 1)待分析的简单语言的词法 * 关键字: begin IF then while do end 所有关键字都是小写. 2)运算符和界符: : = + * - / < <= <> > >= = ( ) # 3)其他单词是标识符(ID)和整数型常数(NUM),通过一下正规式定义: ID=letter (letter|digit)* NUM=digit digit* 4)空格由空白,制表符和换行符组成,空格一般用来分隔ID,NUM,运算符,界符和关键字,此法分析阶段通常被忽略. 3 各种单词符号对应的种别码如表所示
上传时间： 2017-01-08
上传用户：dongqiangqiang
IF you want to it you can download and i m a student,this is a paper,I m wish it can help you.
上传时间： 2014-01-16
上传用户：气温达上千万的
Overview IF you have been wanting to learn Java, check out the newly revised fourth edition of the best-seller Sams Teach Yourself Programming with Java in 24 Hours. This step-by-step tutorial will teach you how to create simple Java programs and applets. Comprised of 24 one-hour lessons, this new edition focuses on key programming concepts and essential Java basics, has been improved by dozens of reader comments, and is reorganized to better cover the latest developments in Java. The book s coverage of core Java programming topics has also been expanded. A great starting point for learning Java, this book is also a great primer to reading sams Teach Yourself Java in 21 Days.
标签： the Overview edition wanting
上传时间： 2017-01-10
上传用户：huyiming139
IF an application works with restricted low level system calls, it must obtain a Microsoft Mobile2Market privileged signature. To get a privileged signature, logo certIFication is now a requirement, not an option! This article shows how to abstract some of the most common issues a developer will encounter when creating a native code application that must be logo certIFied for each platform. windowsmobile5.0以上版本logo注册例子，可以加入自己的工程文件中。
标签： application restricted Microsoft Mobile2Ma
上传时间： 2017-01-16
上传用户：13160677563
javascript动态编程教程源码,IF switch语句
标签： javascript switch IF 动态编程
上传时间： 2017-01-17
上传用户：huangld
PLO源码（c++buider）编译原理课程设计 已经添加了++ += else IF >= <= 等字符
上传时间： 2017-01-18
上传用户：xmsmh
void Knight(int i , int j) { // printf("％d ％dn",i,j) IF (board[i][j] != 0 || i < 0 || i >= Size || j < 0 || j >= Size ) { return } step++ board[i][j]=step IF (step == Size*Size) { showboard() system("PAUSE") return } //DFS Knight(i-2,j-1) //left Knight(i-2,j+1) Knight(i+2,j-1) //right Knight(i+2,j+1) Knight(i-1,j-2) //up Knight(i+1,j-2) Knight(i+1,j+2) //down Knight(i-1,j+2) // board[i][j]=0 step-- }
上传时间： 2014-01-17
上传用户：cxl274287265
int getDivisor(int iNum) { int i = 1 int sum = 0 IF (0 == iNum) { return 1 } while (i <= iNum / 2) { IF (0 == iNum ％ i) { sum++ } i++ } return (sum+1) }
标签： int iNum getDivisor return
上传时间： 2013-12-17
上传用户：frank1234
that main is usb and chuankou s contact and are used,I don t konw IF they are used
上传时间： 2017-01-24
上传用户：wangdean1101
摘要 ： 介绍 丁 DSP技 术 中的双音 多频 (DTMF)技术 ．以及产生 与检测 Drl?,IF信 号 的几 种方 法 ．提 出并 详细 推导 了利 用 Gt~rtze]算法实现滤波器组的方法 及利用仿真软件对 DTMF进芋亍模拟设计的过程 ．
上传时间： 2013-12-22
上传用户：leehom61
