标签：string std 成功 pre end 递归下降 for stdio.h token
递归下降分析：
对于给定的文法G[E] :
E→E+T|E-T|T T→T*F| T/F|F F→(E)|i
消除左递归后的文法是： E→TE‘
E‘→+TE‘|-TE‘|∑
T→FT‘
T‘→*FT‘|/FT‘|∑
F→(E)|i
是否是LL(1)文法？
select(E→TE‘)=first(TE‘)={(,i} select(E‘→+TE‘)=first(+TE‘)={+} select(E‘→-TE‘)=first(-TE‘)={-} select(E‘→∑)=follow(E‘)={),#} select(T→FT‘)=first(FT‘)={(,i} select(T‘→*FT‘)=first(*FT‘)={*} select(T‘→/FT‘)=first(/FT‘)={/} select(T‘→∑)=follow(T‘)={+,-,),#) select(F→(E))=first((E))={(} select(F→i)=first(i)={i}
由上分析，得知此文法满足LL（1）文法
代码如下：
#include <stdio.h> #include <string.h> char prog[800], token[20]; char ch; int syn,p,m,n,sum; char * ch[6]= {"begin","if","then","while","do","end"}; main() { p=0; printf("\n 请输入语句(以#结束):\n"); do{ ch=getchar(); prog[p++]=ch; }while (ch!=‘#‘); p=0; do{ scaner(); switch(syn) { case 11: printf("\n(%d,%d)",syn,sum); break; case -1: printf("\n(%s,)",token);break; default: printf("\n(%d,%s)",syn, token); } }while (syn!=0); p=0; scaner(); E(); if (syn==0) printf("\n 语法成功.\n"); else printf("\n 语法失败 \n"); } scaner() { for (n=0;n<20;n++) token[n]=NULL; m=0; sum=0; ch=prog[p++]; while (ch==‘ ‘) {ch=prog[p++];} if (ch>=‘a‘&& ch<=‘z‘) {while (ch>=‘a‘&& ch<=‘z‘||ch>=‘0‘ && ch<=‘9‘) { token[m++]=ch; ch=prog[p++]; } syn=10;p--; for (n=0;n<6;n++) if(strcmp(token,ch[n])==0) {syn=n+1;break;} } else if(ch>=‘0‘ && ch<=‘9‘) {while (ch>=‘0‘ && ch<=‘9‘) {sum=sum*10+(ch-‘0‘); ch=prog[p++];} syn=11;p--; } else switch(ch) { case ‘<‘: token[m++]=ch; ch=prog[p++]; if (ch==‘>‘) {syn=21;token[m++]=ch;} else if (ch==‘=‘) {syn=22;token[m++]=ch;} else {syn=20;p--;} break; case ‘>‘: m=0; token[m++]=ch; ch=prog[p++]; if (ch==‘=‘){syn=24;token[m++]=ch;} else {syn=23;p--;} break; case ‘:‘: m=0; token[m++]=ch; ch=prog[p++]; if (ch==‘=‘){syn=18;token[m++]=ch;} else {syn=17;p--;} break; case ‘+‘: syn=13;token[0]=ch;break; case ‘-‘: syn=14;token[0]=ch;break; case ‘*‘: syn=15;token[0]=ch;break; case ‘/‘: syn=16;token[0]=ch;break; case ‘=‘: syn=25;token[0]=ch;break; case ‘;‘: syn=26;token[0]=ch;break; case ‘(‘: syn=27;token[0]=ch;break; case ‘)‘: syn=28;token[0]=ch;break; case ‘#‘: syn=0; token[0]=ch;break; default: syn=-1;token[0]=ch; } } E() {printf("E "); T(); E1(); } E1() {printf("E1 "); if (syn==13) { scaner(); T(); E1(); } else { if (syn!=28 && syn!=0) error(); } } T() {printf("T "); F(); T1(); } T1() {printf("T1 "); if (syn==15) { scaner(); F(); T1(); } else { if (syn!=28 && syn!=0 && syn!=13) error(); } } F() {printf("F "); if (syn==27) { scaner(); E(); if(syn==28) scaner(); else error(); } else if (syn==11 || syn==10) scaner(); } error() { printf("\n (%d,%s)语法错误! \n",syn, token); }
标签：string std 成功 pre end 递归下降 for stdio.h token
原文地址：http://www.cnblogs.com/cgq520/p/6188742.html
