詞法分析 1 試驗目的 設計,編制并調試一個此法分析程序,加深對此法分原理的理解. 2 試驗要求 1)待分析的簡單語言的詞法 * 關鍵字: begin if then while do end 所有關鍵字都是小寫. 2)運算符和界符: : = + * - / < <= <> > >= = ( ) # 3)其他單詞是標識符(ID)和整數型常數(NUM),通過一下正規式定義: ID=letter (letter|digit)* NUM=digit digit* 4)空格由空白,制表符和換行符組成,空格一般用來分隔ID,NUM,運算符,界符和關鍵字,此法分析階段通常被忽略. 3 各種單詞符號對應的種別碼如表所示
上傳時間： 2017-01-08
上傳用戶：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.
上傳時間： 2014-01-16
上傳用戶：氣溫達上千萬的
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.
標簽： the Overview edition wanting
上傳時間： 2017-01-10
上傳用戶：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注冊例子，可以加入自己的工程文件中。
標簽： application restricted Microsoft Mobile2Ma
上傳時間： 2017-01-16
上傳用戶：13160677563
javascript動態編程教程源碼,if switch語句
標簽： javascript switch if 動態編程
上傳時間： 2017-01-17
上傳用戶：huangld
PLO源碼（c++buider）編譯原理課程設計 已經添加了++ += else if >= <= 等字符
上傳時間： 2017-01-18
上傳用戶：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-- }
上傳時間： 2014-01-17
上傳用戶：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) }
標簽： int iNum getDivisor return
上傳時間： 2013-12-17
上傳用戶：frank1234
that main is usb and chuankou s contact and are used,I don t konw if they are used
上傳時間： 2017-01-24
上傳用戶：wangdean1101
摘要 ： 介紹 丁 DSP技 術 中的雙音 多頻 (DTMF)技術 ．以及產生 與檢測 Drl?,if信 號 的幾 種方 法 ．提 出并 詳細 推導 了利 用 Gt~rtze]算法實現濾波器組的方法 及利用仿真軟件對 DTMF進芋亍模擬設計的過程 ．
上傳時間： 2013-12-22
上傳用戶：leehom61
