introduces a serial LDPC encoder and decoder of C forms prepared environment. Ne...

比较简单的一个词法分析器，帮助学编译原理的初学者理解，可作为教学用途-relatively simple one lexical analyzer to help the school build understanding of the principle of beginners can use as a teaching

一个C语言编译器源代码，外国人写的东东，生成目标代码时需要有masm配合。-a C language compiler source code, written in the Eastern foreigners, generating object code when it can meet the need.

这是编译原理里面用来判断输入的东西是是否是符合数学运算公式-This is the compiler used to determine the principle of which is whether the importation of things are in line with the mathematical formula

Pascal 到C的转换器 -Converter from Pascal to C

一个用C++编写的C语言词法分析程序，本人在做WEBSPHERE MQ的收发信息时需要对远端发过来的数据进行解析，然后才能确定要发送到哪一个终端，因此写了对词法分析做了一定的研究，也就写了这么一个程序.-prepared by the C language lexical analysis procedures, I do WebSphere MQ the need to send and receive information sent to the remote data analysis, in order to determine where to send a terminal, wrote a pair of lexical analysis done some research, it was such a procedure.

一种视频流H264压缩加密方法

编译原理实验中的yacc源码，是一个基于小类C语言的yacc程序，有助于学习yacc的编写。-principle experiment compiler generates the source code, is based on a subset of the C language yacc procedures, contribute to the preparation of learning yacc.

本程序不能编译除大小写字母之外的编码 本程序需事先建立字母权值文后件才可运行 本程序从文件中读入文字，编译后写入文件，所以屏幕上不会直接出现答案-not build this program in addition to case-insensitive outside of the coding process required prior to the establishment of the letters after the value of the text pieces can Win OK this program read from the file into the text, compiled into the paper, not directly on the screen appeared the answer

1．词法分析 设计、编制并调试该C语言子集的词法分析程序，要求结果以二元组形式保存在文件中。-1. Lexical analysis design, develop and debug the C language subset of the lexical analysis procedure that requires results to binary form group to preserve the documents.

词法分析程序，可对以下的C源程序进行分析：main() {int a[12] ,sum for(i=1 i=1 j--){ if(i==j&&i+j==13)sum+=a[i][j] } } printf("%c",sum) }-lexical analysis procedures, right below the C source code analysis : main () (int a [12] [12], the sum for (i = 1 ilt; I = 12) (for (j = 1 JLT; J = 12) Scanf ( "% d", a [i] [j])) for (i = 12 IGT; a = i--) (for (j = 12 IGT; = 1 j--) (if (i == ji j == 13) sum = a [i] [j])) printf ( "% c", sum