small c for dos/z80

这是一个用c++写的预测分析法程序,是编译原理课程设计的内容,可以预测分析语言-c write with the forecast analysis procedures, compiler theory of curriculum design, analysis can predict language

#include "define.h" #include #include #include #include FILE *fp /*////////////////////////////////////////////////////////////////////////////// This is a pretreatment. /////////////////////////////////////////////////////////////////////////////*/ void readIntoBuffer(char buffer[256]) { char ch1 char temp int i for(i=0 i

该程序能够实现简单的c语言词法，语法，语义分析。- This procedure can realize the simple c language morphology, the grammar, the semantic analysis.

语义分析的功能： 第一，审查每个语法结构的静态语义，即验证语法结构合法的程序是否真正有意义。有时把这个工作称为静态语义分析或静态审查。第二，如果静态语义正确，语义处理则要执行真正的翻译，即，或者将源程序翻译成程序的一种中间表示形式（中间代码），或者将源程序翻译成目标代码。 有报告-Semantic analysis of the functions: First, review each of the grammatical structure of the static semantics, that is, the grammatical structure to verify whether the procedures legally meaningful. Sometimes this is called static or static semantic analysis review. Secondly, if the correct static semantics, semantic processing should be the implementation of a genuine translation, that is, or will be translated into procedures source an intermediate representation (intermediate code), or source code translated into object code. There are reports

编译器,词法分析和语法分析,里面利用到了哈希表来加快搜索字符表速度.-compiler, lexical analysis and parsing, to use inside the hash table to expedite the search speed character tables.

简单basic语言编译器-simple basic language compiler

这是学了编译原理后在老师的帮助下，编的一个集词法，语法，语义于一体的分析其源码！可直接在VC6.0环境下运行。价值很高啊！ -This was learned after compiling principle teacher

数控系统中的解释器源代码，可以独立的运行，直接在dos环境下解释G代码-NC system for the interpretation of the source code, independent of the operation, directly in the environment explained dos G code

java文件注释过滤器,将文件中的注释过滤掉.-java document filters, the document notes strip.

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