实验一 词法分析器的设计一、实验目的 （1）学会针对转换图实现相应的高级语言源程序。 （2）深刻领会状态转换图的含义，逐步理解有限自动机。二、实验内容 （1）某计算机语言的编译程序的词法分析部分实现。 （2）从左到右扫描每行该语言源程序的符号，拼成单词，换成统一的内部表示（token），送给语法分析程序。三、实现原理程序中先判断这个句语句中每个单元为关键字、常数、运算符、界符，对与不同的单词符号给出不同编码形式的编码，用以区分之。PL/0语言的EBNF表示：=;：=|;:=+|-:=*|/:=|#|=:=a|b|X|Y|Z:=0|1|2|8|9四、设计过程1 关键字：void，main，if，then，break，int，Char，float，include，for，while，printfscanf 并为小写。 2+”;”-”;”*”;”/”;”:=“;”:”;”“;”“;”=“;”“;”=“;”(“;”)”;”;”;”#”为运算符。3 其他标记 如字符串，表示以字母开头的标识符。 4 空格符跳过。 5 各符号对应种别码 关键字分别对应1-13运算符分别对应401-418，501-513。字符串对应100常量对应200结束符#五、心得体会其实匹配并不困难，主要是C+知识要求相对较高，只要把握住指针就好了。附源程序：#include#include#include #includeint i,j,k,flag,number,status;/*status which is use to judge the string is keywords or not!*/char ch;char words10 = ;char program500;int Scan(char program) char *keywords13 = void,main,if,then,break,int, char,float,include,for,while,printf, scanf;number = 0;status = 0;j = 0;ch = programi+; /* To handle the lettle space ands tab*/ /*handle letters*/if (ch = a) & (ch = a) & (ch = z )wordsj+=ch;ch=programi+; i-; wordsj+ = 0; for (k = 0; k = 0) & (ch = 0 ) & (ch : if (ch = ) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch; wordsj = 0; flag = 403; else i-; flag = 404; break;case: if (ch = ) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch; wordsj = 0; flag = 405; else i-; flag = 406; break;case!: if (ch = !) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch; wordsj = 0; flag = 407; else i-; flag = 408; break;case+: if (ch = +) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch; wordsj = 0; flag = 409; else if (ch = +) wordsj+ = ch; wordsj = 0; flag = 410; else i-;flag = 411; break;case-: if (ch = -) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch;wordsj = 0;flag = 412; else if( ch = -) wordsj+ = ch;wordsj = 0;flag = 413; else i-;flag = 414; break;case*: if (ch = *) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch; wordsj = 0; flag = 415; else i-; flag = 416; break; case/: if (ch = /) wordsj+ = ch; wordsj = 0; ch = programi+; if (ch = =) wordsj+ = ch; wordsj = 0; flag = 417; else i-;