編譯原理詞法分析和語法分析報告代碼(C語言版)

1、 詞法分析一、 實驗?zāi)康脑O(shè)計、編制并調(diào)試一個詞法分析程序，加深對詞法分析原理的理解。二、 實驗要求 詞法分析程序的功能：輸入：所給文法的源程序字符串。輸出：二元組（syn,token或sum）構(gòu)成的序列。其中：syn為單詞種別碼； token為存放的單詞自身字符串； sum為整型常數(shù)。三、詞法分析程序的C語言程序源代碼：#include <stdio.h>#include <string.h>char prog80,token8,ch;int syn,p,m,n,sum;char *rwtab6="begin","if",&quo

2、t;then","while","do","end" scaner();main()p=0; printf("n please input a string(end with '#'):/n"); do scanf("%c",&ch); progp+=ch; while(ch!='#'); p=0; do scaner(); switch(syn) case 11:printf("( %-10d%5d )n",sum,syn)

3、; break; case -1:printf("you have input a wrong stringn"); getch(); exit(0); default: printf("( %-10s%5d )n",token,syn); break; while(syn!=0); getch(); scaner() sum=0; for(m=0;m<8;m+)tokenm+=NULL; ch=progp+; m=0; while(ch=' ')|(ch='n')ch=progp+; if(ch<='

4、z')&&(ch>='a')|(ch<='Z')&&(ch>='A') while(ch<='z')&&(ch>='a')|(ch<='Z')&&(ch>='A')|(ch>='0')&&(ch<='9') tokenm+=ch; ch=progp+; p-; syn=10; for(n=0;n<6;n+)

5、 if(strcmp(token,rwtabn)=0) syn=n+1; break; else if(ch>='0')&&(ch<='9') while(ch>='0')&&(ch<='9') sum=sum*10+ch-'0' ch=progp+; p-; syn=11; else switch(ch) case '<':tokenm+=ch; ch=progp+; if(ch='=') syn=22; tokenm+

6、=ch; else syn=20; p-; break; case '>':tokenm+=ch; ch=progp+; if(ch='=') syn=24; tokenm+=ch; else syn=23; p-; break; case '+': tokenm+=ch; ch=progp+; if(ch='+') syn=17; tokenm+=ch; else syn=13; p-; break; case '-':tokenm+=ch; ch=progp+; if(ch='-') sy

7、n=29; tokenm+=ch; else syn=14; p-; break; case '!':ch=progp+; if(ch='=') syn=21; tokenm+=ch; else syn=31; p-; break; case '=':tokenm+=ch; ch=progp+; if(ch='=') syn=25; tokenm+=ch; else syn=18; p-; break; case '*': syn=15; tokenm+=ch; break; case '/': s

8、yn=16; tokenm+=ch; break; case '(': syn=27; tokenm+=ch; break; case ')': syn=28; tokenm+=ch; break; case '': syn=5; tokenm+=ch; break; case '': syn=6; tokenm+=ch; break; case '': syn=26; tokenm+=ch; break; case '"': syn=30; tokenm+=ch; break; case '#': syn=0; tokenm+=ch; break; case ':':syn=17; tokenm+=ch; break; default: syn=-1; break; tokenm+='0' 五、結(jié)果分析：輸入begin x:=9: if x>9 then x:=2*x+1/3; end # 后經(jīng)詞法分析輸出如下序列：(begin 1)(x 10)(：17)(= 18)(9 11)(；2