第五次作业----词法分析程序的设计与实现

词法分析程序(Lexical Analyzer)要求:

- 从左至右扫描构成源程序的字符流

-  识别出有词法意义的单词(Lexemes

-  返回单词记录(单词类别,单词本身)

-  滤掉空格

-  跳过注释

-  发现词法错误

程序结构:

输入:字符流(什么输入方式,什么数据结构保存)

处理:

–遍历(什么遍历方式)

–词法规则

输出:单词流(什么输出形式)

–二元组

单词类别:

1.标识符(10)

2.无符号数(11)

3.保留字(一词一码)

4.运算符(一词一码)

5.界符(一词一码)

单词符号

种别码

单词符号

种别码

begin

1

:

17

if

2

:=

18

then

3

<

20

while

4

<=

21

do

5

<>

22

end

6

>

23

l(l|d)*

10

>=

24

dd*

11

=

25

+

13

;

26

-

14

(

27

*

15

)

28

/

16

#

0

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//声明变量
char prog[800],dc[8]; //程序段  单词
char ch;  //单词中的字符
int syn,p,m=0; //单词符号种别码   指针p
int n,sum=0;  //整数sum
const char *word[6]={"begin","if","then","while","do","end"};  //保留字
//主函数
int main(void){
    //指针从0开始
    p=0;
    printf("请输入源程序段:");
    //逐个字符读入,放入字符数组prog中,直到'#'停止
    do{
        ch=getchar();
        prog[p++]=ch;   
    }while(ch!='#');
    //指针从0开始
    p=0;
    //逐个字符判断其单词符号种别码
    do{
        //初始化数组dc
        for(n=0;n<8;n++){
            dc[n]=NULL;
        }
        ch=prog[p++];
        //遇到空格指针加1
        while(ch==' '){
            ch=prog[p];
            p++;
        }
        //标识符
        if((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z')){
            m=0;
            while((ch>='0'&&ch<='9')||(ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z')){
                dc[m++]=ch;
                ch=prog[p++];
            }
            p--;
            syn=10;
            //保留字
            for(n=0;n<6;n++){
                if(strcmp(dc,word[n])==0){
                    syn=n+1;
                    break;
                }
            }
        }
        //数字
        else if((ch>='0'&&ch<='9')){
                sum=0;
                while((ch>='0'&&ch<='9')){
                    sum=sum*10+ch-'0';
                    ch=prog[p++];
                }
            p--;
            syn=11;
        }
        else{
            switch(ch){
                case '<':m=0;
                    dc[m++]=ch;
                    ch=prog[p++];
                    if(ch=='>'){
                        syn=22;
                        dc[m++]=ch;
                    }
                    else if(ch=='='){
                        syn=21;
                        dc[m++]=ch;
                    }
                    else{
                        syn=20;
                        p--;
                    }
                    break;
                case '>':m=0;
                    dc[m++]=ch;
                    ch=prog[p++];
                    if(ch=='='){
                        syn=24;
                        dc[m++]=ch;
                    }
                    else{
                        syn=23;
                        p--;
                    }
                    break;
                case ':':m=0;
                    dc[m++]=ch;
                    ch=prog[p++];
                    if(ch=='='){
                        syn=18;
                        dc[m++]=ch;
                    }
                    else{
                        syn=17;
                        p--;
                    }
                    break;
                case '*':syn=15;dc[0]=ch;break;
                case '/':syn=16;dc[0]=ch;break;
                case '+':syn=13;dc[0]=ch;break;
                case '-':syn=14;dc[0]=ch;break;
                case '=':syn=25;dc[0]=ch;break;
                case ';':syn=26;dc[0]=ch;break;
                case '(':syn=27;dc[0]=ch;break;
                case ')':syn=28;dc[0]=ch;break;
                case '#':syn=0;dc[0]=ch;break;
                case ' ':syn=-2;dc[0]=ch;break;       
            }
        }
        switch(syn){
            case 11:printf("(%d,%d) ",syn,sum);break;
            default:printf("(%d,%s) ",syn,dc);break;
        }
    }while(syn!=0);
}

 

原文地址:https://www.cnblogs.com/sgczw/p/11655209.html