1、利用C C+语言求的LR1分析方法如:给定一文法:Z-aAcB|BdA-cDD-aBD|dB-bCC-BcA|能求出其first集: First(Z)=a,b First(A)=c First(B)=b First(C)=b, First(D)=a,d First(a)=a First(b)=b First(c)=c First(d)=d first()=follow集: FOLLOW(A) = c, #, d, a FOLLOW(B) = #, d, a, c FOLLOW(C) = #, d, a, c FOLLOW(D) = c, #, d, a FOLLOW(Z) = # select
2、集: SELECT(Z-aAcB) = a SELECT(Z-Bd) = b SELECT(A-cD) = c SELECT(D-aBD) = a SELECT(D-d) = d SELECT(B-bC) = b SELECT(C-BcA) = b SELECT(C-) = ,#,d,a,c给定一字符处让其判定如acabdcb,结果能接受程序:#include#include#include/*/int count=0; /*分解的产生式的个数*/int number; /*所有终结符和非终结符的总数*/char start; /*开始符号*/char termin50; /*终结符号*/ch
3、ar non_ter50; /*非终结符号*/char v50; /*所有符号*/char left50; /*左部*/char right5050; /*右部*/char first5050,follow5050; /*各产生式右部的FIRST和左部的FOLLOW集合*/char first15050; /*所有单个符号的FIRST集合*/char select5050; /*各单个产生式的SELECT集合*/char f50,F50; /*记录各符号的FIRST和FOLLOW是否已求过*/char empty20; /*记录可直接推出的符号*/char TEMP50; /*求FOLLOW时
4、存放某一符号串的FIRST集合*/int validity=1; /*表示输入文法是否有效*/int ll=1; /*表示输入文法是否为LL(1)文法*/int M2020; /*分析表*/char choose; /*用户输入时使用*/char empt20; /*求_emp()时使用*/char fo20; /*求FOLLOW集合时使用*/*判断一个字符是否在指定字符串中*/int in(char c,char *p)int i;if(strlen(p)=0)return(0);for(i=0;i+) if(pi=c) return(1); /*若在,返回1*/if(i=strlen(p)
5、 return(0); /*若不在,返回0*/*得到一个不是非终结符的符号*/char c()char c=A; while(in(c,non_ter)=1)c+;return(c);/*分解含有左递归的产生式*/void recur(char *point) /*完整的产生式在point中*/ int j,m=0,n=3,k;char temp20,ch;ch=c(); /*得到一个非终结符*/k=strlen(non_ter);non_terk=ch;non_terk+1=0;for(j=0;j=strlen(point)-1;j+) if(pointn=point0) /*如果|后的首符
6、号和左部相同*/ for(j=n+1;j=strlen(point)-1;j+) while(pointj!=|&pointj!=0) tempm+=pointj+; leftcount=ch; memcpy(rightcount,temp,m); rightcountm=ch; rightcountm+1=0; m=0; count+; if(pointj=|) n=j+1; break; else /*如果|后的首符号和左部不同*/ leftcount=ch; rightcount0=; rightcount1=0; count+; for(j=n;j=strlen(point)-1;j+
7、) if(pointj!=|) tempm+=pointj; else leftcount=point0; memcpy(rightcount,temp,m); rightcountm=ch; rightcountm+1=0; printf( count=%d ,count); m=0; count+; leftcount=point0; memcpy(rightcount,temp,m); rightcountm=ch; rightcountm+1=0; count+; m=0;/*分解不含有左递归的产生式*/void non_re(char *point) int m=0,j;char t
8、emp20;for(j=3;j=strlen(point)-1;j+) if(pointj!=|) tempm+=pointj;else leftcount=point0; memcpy(rightcount,temp,m); rightcountm=0; m=0; count+; leftcount=point0; memcpy(rightcount,temp,m); rightcountm=0; count+;m=0;/*读入一个文法*/char grammer(char *t,char *n,char *left,char right5050)char vn50,vt50;char s;
9、char p5050;int i,j,k;printf(请输入文法的非终结符号串:); scanf(%s,vn);getchar(); i=strlen(vn); memcpy(n,vn,i);ni=0;printf(请输入文法的终结符号串:); scanf(%s,vt);getchar(); i=strlen(vt); memcpy(t,vt,i);ti=0; printf(请输入文法的开始符号:);scanf(%c,&s);getchar();printf(请输入文法产生式的条数:); scanf(%d,&i);getchar(); for(j=1;j=i;j+)printf(请输入文法的
10、第%d条(共%d条)产生式:,j,i);scanf(%s,pj-1); getchar(); for(j=0;j) printf(ninput error!); validity=0; return(0); /*检测输入错误*/ for(k=0;k=i-1;k+) /*分解输入的各产生式*/ if(pk3=pk0) recur(pk);else non_re(pk);return(s);/*将单个符号或符号串并入另一符号串*/void merge(char *d,char *s,int type) /*d是目标符号串,s是源串,type1,源串中的 一并并入目串; type2,源串中的 不并入
11、目串*/ int i,j;for(i=0;i=strlen(s)-1;i+) if(type=2&si=) ;else for(j=0;j+) if(jstrlen(d)&si=dj) break; if(j=strlen(d) dj=si; dj+1=0; break; /*求所有能直接推出的符号*/void emp(char c) /*即求所有由 推出的符号*/char temp10;int i;for(i=0;i=count-1;i+)if(righti0=c&strlen(righti)=1) temp0=lefti; temp1=0; merge(empty,temp,1); emp
12、(lefti);/*求某一符号能否推出 */int _emp(char c) /*若能推出,返回1;否则,返回0*/int i,j,k,result=1,mark=0;char temp20;temp0=c;temp1=0;merge(empt,temp,1);if(in(c,empty)=1)return(1);for(i=0;i+)if(i=count) return(0);if(lefti=c) /*找一个左部为c的产生式*/ j=strlen(righti); /*j为右部的长度*/ if(j=1&in(righti0,empty)=1) return(1); else if(j=1&
13、in(righti0,termin)=1) return(0); else for(k=0;k=j-1;k+) if(in(rightik,empt)=1) mark=1; if(mark=1) continue; else for(k=0;k=j-1;k+) result*=_emp(rightik); temp0=rightik; temp1=0; merge(empt,temp,1); if(result=0&icount) continue; else if(result=1&icount) return(1);/*判断读入的文法是否正确*/int judge() int i,j;fo
14、r(i=0;i=count-1;i+)if(in(lefti,non_ter)=0) /*若左部不在非终结符中,报错*/ printf(nerror1!); validity=0; return(0);for(j=0;j=strlen(righti)-1;j+) if(in(rightij,non_ter)=0&in(rightij,termin)=0&rightij!=) /*若右部某一符号不在非终结符、终结符中且不为 ,报错*/ printf(nerror2!); validity=0; return(0); return(1);/*求单个符号的FIRST*/void first2(int
15、 i) /*i为符号在所有输入符号中的序号*/ char c,temp20;int j,k,m;c=vi;char ch=;emp(ch);if(in(c,termin)=1) /*若为终结符*/ first1i0=c; first1i1=0; else if(in(c,non_ter)=1) /*若为非终结符*/for(j=0;j=count-1;j+) if(leftj=c) if(in(rightj0,termin)=1|rightj0=) temp0=rightj0; temp1=0; merge(first1i,temp,1); else if(in(rightj0,non_ter)
16、=1) if(rightj0=c) continue; for(k=0;k+) if(vk=rightj0) break; if(fk=0) first2(k); fk=1; merge(first1i,first1k,2); for(k=0;k=strlen(rightj)-1;k+) empt0=0; if(_emp(rightjk)=1&k=0) firsti0=; firsti1=0; else TEMP0=; TEMP1=0; else for(j=0;j+) if(vj=p0) break; if(i=0) memcpy(firsti,first1j,strlen(first1j)
17、; firstistrlen(first1j)=0; else memcpy(TEMP,first1j,strlen(first1j); TEMPstrlen(first1j)=0; else /*如果右部为符号串*/for(j=0;j+) if(vj=p0) break;if(i=0) merge(firsti,first1j,2);else merge(TEMP,first1j,2);for(k=0;k=length-1;k+) empt0=0; if(_emp(pk)=1&k=0) merge(firsti,first1m,2); else merge(TEMP,first1m,2);
18、else if(_emp(pk)=1&k=length-1) temp0=; temp1=0; if(i=0) merge(firsti,temp,1); else merge(TEMP,temp,1); else if(_emp(pk)=0) break;/*求各产生式左部的FOLLOW*/void FOLLOW(int i)int j,k,m,n,result=1;char c,temp20;c=non_teri; /*c为待求的非终结符*/temp0=c;temp1=0;merge(fo,temp,1);if(c=start) /*若为开始符号*/temp0=#;temp1=0;merg
19、e(followi,temp,1); for(j=0;j=count-1;j+)if(in(c,rightj)=1) /*找一个右部含有c的产生式*/ for(k=0;k+) if(rightjk=c) break; /*k为c在该产生式右部的序号*/ for(m=0;m+) if(vm=leftj) break; /*m为产生式左部非终结符在所有符号中的序号*/ if(k=strlen(rightj)-1) /*如果c在产生式右部的最后*/ if(in(vm,fo)=1) merge(followi,followm,1); continue; if(Fm=0) FOLLOW(m); Fm=1
20、; merge(followi,followm,1); else /*如果c不在产生式右部的最后*/ for(n=k+1;n=strlen(rightj)-1;n+) empt0=0; result*=_emp(rightjn); if(result=1) /*如果右部c后面的符号串能推出*/ if(in(vm,fo)=1) /*避免循环递归*/ merge(followi,followm,1); continue; if(Fm=0) FOLLOW(m); Fm=1; merge(followi,followm,1); for(n=k+1;n=strlen(rightj)-1;n+) tempn-k-1=rightjn; tempstrlen(rightj)-k-1=0; FIRST(-1,temp); merge(followi,TEMP,2); Fi=1;/*判断读入文法是否为一个LL(1)文法*
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