LL語(yǔ)法分析程序

1、 編譯原理上機(jī)實(shí)驗(yàn)報(bào)告題目：LL(1)語(yǔ)法分析程序1.設(shè)計(jì)要求 （1）對(duì)輸入文法，它能判斷是否為L(zhǎng)L(1)文法，若是，則轉(zhuǎn)（2）；否則報(bào)錯(cuò)并終止； （2）輸入已知文法，由程序自動(dòng)生成它的LL(1)分析表； （3）對(duì)于給定的輸入串，應(yīng)能判斷識(shí)別該串是否為給定文法的句型。2.分析 該程序可分為如下幾步：（1）讀入文法 （2）判斷正誤 （3）若無誤，判斷是否為L(zhǎng)L(1)文法 （4）若是，構(gòu)造分析表；（5）由總控算法判斷輸入符號(hào)串是否為該文法的句型。有效？讀入文法開始3.流程圖是是LL(1)文法？結(jié)束報(bào)錯(cuò)判斷句型是4.源程序/* LL1語(yǔ)法分析程序*/#include#include#include/

2、*/int count=0; /*分解的產(chǎn)生式的個(gè)數(shù)*/int number; /*所有終結(jié)符和非終結(jié)符的總數(shù)*/char start; /*開始符號(hào)*/char termin50; /*終結(jié)符號(hào)*/char non_ter50; /*非終結(jié)符號(hào)*/char v50; /*所有符號(hào)*/char left50; /*左部*/char right5050; /*右部*/char first5050,follow5050; /*各產(chǎn)生式右部的FIRST和左部的FOLLOW集合*/char first15050; /*所有單個(gè)符號(hào)的FIRST集合*/char select5050; /*各單個(gè)產(chǎn)生式的

3、SELECT集合*/char f50,F50; /*記錄各符號(hào)的FIRST和FOLLOW是否已求過*/char empty20; /*記錄可直接推出的符號(hào)*/char TEMP50; /*求FOLLOW時(shí)存放某一符號(hào)串的FIRST集合*/int validity=1; /*表示輸入文法是否有效*/int ll=1; /*表示輸入文法是否為L(zhǎng)L(1)文法*/int M2020; /*分析表*/char choose; /*用戶輸入時(shí)使用*/char empt20; /*求_emp()時(shí)使用*/char fo20; /*求FOLLOW集合時(shí)使用*/* 判斷一個(gè)字符是否在指定字符串中*/int in

4、(char c,char *p)/int i;size_t 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) return(0); /*若不在，返回0*/* 得到一個(gè)不是非終結(jié)符的符號(hào)*/char c()char c=A; while(in(c,non_ter)=1)c+;return(c);/* 分解含有左遞歸的產(chǎn)生式*/void recur(char *point) /*完整的產(chǎn)生式在point中*/ int j,m=0,n=3,k;char temp20,ch;ch=c();

5、 /*得到一個(gè)非終結(jié)符*/k=strlen(non_ter);non_terk=ch;non_terk+1=0;for(j=0;size_t(j)=strlen(point)-1;j+)if(pointn=point0) /*如果|后的首符號(hào)和左部相同*/for(j=n+1;size_t(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

6、=|)n=j+1;break;else /*如果|后的首符號(hào)和左部不同*/leftcount=ch;rightcount0=;rightcount1=0;count+;for(j=n;size_t(j)=strlen(point)-1;j+) 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(righ

7、tcount,temp,m); rightcountm=ch; rightcountm+1=0;count+; m=0;/* 分解不含有左遞歸的產(chǎn)生式*/void non_re(char *point) int m=0,j;char temp20;for(j=3;size_t(j)=strlen(point)-1;j+) if(pointj!=|) tempm+=pointj;else leftcount=point0; memcpy(rightcount,temp,m); rightcountm=0;m=0;count+; leftcount=point0; memcpy(rightcoun

8、t,temp,m); rightcountm=0; count+;m=0;/* 讀入一個(gè)文法*/char grammer(char *t,char *n,char *left,char right5050)char vn50,vt50;char s;char p5050;int i,j,k;printf(n請(qǐng)輸入文法的非終結(jié)符號(hào)串：); scanf(%s,vn);getchar(); i=strlen(vn); memcpy(n,vn,i);ni=0;printf(請(qǐng)輸入文法的終結(jié)符號(hào)串：); scanf(%s,vt);getchar(); i=strlen(vt); memcpy(t,vt,

9、i);ti=0; printf(請(qǐng)輸入文法的開始符號(hào)：);scanf(%c,&s);getchar();printf(請(qǐng)輸入文法產(chǎn)生式的條數(shù)：); scanf(%d,&i);getchar(); for(j=1;j=i;j+)printf(請(qǐng)輸入文法的第%d條（共%d條）產(chǎn)生式：,j,i);scanf(%s,pj-1); getchar(); for(j=0;j)printf(ninput error!); validity=0;return(0); /*檢測(cè)輸入錯(cuò)誤*/ for(k=0;k=i-1;k+) /*分解輸入的各產(chǎn)生式*/ if(pk3=pk0) recur(pk);else no

10、n_re(pk);return(s);/* 將單個(gè)符號(hào)或符號(hào)串并入另一符號(hào)串*/void merge(char *d,char *s,int type) /*d是目標(biāo)符號(hào)串，s是源串，type1，源串中的 一并并入目串； type2，源串中的 不并入目串*/ int i,j;for(i=0;size_t(i)=strlen(s)-1;i+) if(type=2&si=);elsefor(j=0;j+) if(size_t(j)strlen(d)&si=dj) break; if(size_t(j)=strlen(d) dj=si; dj+1=0; break;/* 求所有能直接推出的符號(hào)*/v

11、oid emp(char c) /*即求所有由 推出的符號(hào)*/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(lefti);/* 求某一符號(hào)能否推出 */int _emp(char c) /*若能推出，返回1；否則，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

12、);for(i=0;i+)if(i=count) return(0);if(lefti=c) /*找一個(gè)左部為c的產(chǎn)生式*/ j=strlen(righti); /*j為右部的長(zhǎng)度*/if(j=1&in(righti0,empty)=1) return(1);else if(j=1&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=righti

13、k;temp1=0;merge(empt,temp,1); if(result=0&icount) continue; else if(result=1&icount) return(1);/* 判斷讀入的文法是否正確*/int judge() int i,j;for(i=0;i=count-1;i+)if(in(lefti,non_ter)=0) /*若左部不在非終結(jié)符中，報(bào)錯(cuò)*/printf(nerror1!);validity=0;return(0);for(j=0;size_t(j)=strlen(righti)-1;j+)if(in(rightij,non_ter)=0&in(rig

14、htij,termin)=0&rightij!=) /*若右部某一符號(hào)不在非終結(jié)符、終結(jié)符中且不為 ，報(bào)錯(cuò)*/printf(nerror2!);validity=0;return(0);return(1);/* 求單個(gè)符號(hào)的FIRST*/void first2(int i) /*i為符號(hào)在所有輸入符號(hào)中的序號(hào)*/ char c,temp20;int j,k,m;c=vi;char ch=;emp(ch);if(in(c,termin)=1) /*若為終結(jié)符*/ first1i0=c; first1i1=0; else if(in(c,non_ter)=1) /*若為非終結(jié)符*/for(j=0;

15、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)=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;size_t(k)=strlen(rightj)-1;k+)empt0=0;if(_emp(rightjk

16、)=1&size_t(k)=0) firsti0=; firsti1=0;elseTEMP0=;TEMP1=0;elsefor(j=0;j+)if(vj=p0)break;if(i=0) memcpy(firsti,first1j,strlen(first1j); firstistrlen(first1j)=0;elsememcpy(TEMP,first1j,strlen(first1j);TEMPstrlen(first1j)=0; else /*如果右部為符號(hào)串*/for(j=0;j+)if(vj=p0)break;if(i=0) merge(firsti,first1j,2);elsem

17、erge(TEMP,first1j,2);for(k=0;k=length-1;k+)empt0=0;if(_emp(pk)=1&k=0) merge(firsti,first1m,2);elsemerge(TEMP,first1m,2); else if(_emp(pk)=1&k=length-1) temp0=;temp1=0;if(i=0) merge(firsti,temp,1); elsemerge(TEMP,temp,1);else if(_emp(pk)=0)break;/* 求各產(chǎn)生式左部的FOLLOW*/void FOLLOW(int i)int j,k,m,n,result

18、=1;char c,temp20;c=non_teri; /*c為待求的非終結(jié)符*/temp0=c;temp1=0;merge(fo,temp,1);if(c=start) /*若為開始符號(hào)*/temp0=#;temp1=0;merge(followi,temp,1); for(j=0;j=count-1;j+)if(in(c,rightj)=1) /*找一個(gè)右部含有c的產(chǎn)生式*/for(k=0;k+)if(rightjk=c)break; /*k為c在該產(chǎn)生式右部的序號(hào)*/ for(m=0;m+)if(vm=leftj)break; /*m為產(chǎn)生式左部非終結(jié)符在所有符號(hào)中的序號(hào)*/if(si

19、ze_t(k)=strlen(rightj)-1) /*如果c在產(chǎn)生式右部的最后*/if(in(vm,fo)=1)merge(followi,followm,1);continue; if(Fm=0)FOLLOW(m);Fm=1;merge(followi,followm,1);else /*如果c不在產(chǎn)生式右部的最后*/for(n=k+1;size_t(n)=strlen(rightj)-1;n+)empt0=0;result*=_emp(rightjn);if(result=1) /*如果右部c后面的符號(hào)串能推出*/ if(in(vm,fo)=1) /*避免循環(huán)遞歸*/merge(foll

20、owi,followm,1);continue;if(Fm=0) FOLLOW(m); Fm=1; merge(followi,followm,1);for(n=k+1;size_t(n)=strlen(rightj)-1;n+) tempn-k-1=rightjn; tempstrlen(rightj)-k-1=0;FIRST(-1,temp);merge(followi,TEMP,2);Fi=1;/* 判斷讀入文法是否為一個(gè)LL(1)文法*/int ll1() int i,j,length,result=1;char temp50;for(j=0;j=49;j+) /*初始化*/first

21、j0=0; followj0=0;first1j0=0;selectj0=0;TEMPj=0;tempj=0;fj=0;Fj=0;for(j=0;size_t(j)=strlen(v)-1;j+) first2(j); /*求單個(gè)符號(hào)的FIRST集合*/printf(nfirst1:);for(j=0;size_t(j)=strlen(v)-1;j+)printf(%c:%s ,vj,first1j); printf(nempty:%s,empty);printf(n:n_emp:);for(j=0;size_t(j)=strlen(v)-1;j+) printf(%d ,_emp(vj);f

22、or(i=0;i=count-1;i+) FIRST(i,righti); /*求FIRST*/printf(n);for(j=0;size_t(j)=strlen(non_ter)-1;j+) /*求FOLLOW*/if(foj=0)fo0=0; FOLLOW(j); printf(nfirst:);for(i=0;i=count-1;i+) printf(%s ,firsti);printf(nfollow:); for(i=0;size_t(i)=strlen(non_ter)-1;i+) printf(%s ,followi);for(i=0;i=count-1;i+) /*求每一產(chǎn)生

23、式的SELECT集合*/ memcpy(selecti,firsti,strlen(firsti); selectistrlen(firsti)=0;for(j=0;size_t(j)=strlen(righti)-1;j+)result*=_emp(rightij);if(strlen(righti)=1&righti0=)result=1;if(result=1)for(j=0;j+)if(vj=lefti)break;merge(selecti,followj,1);printf(nselect:);for(i=0;i=count-1;i+) printf(%s ,selecti);me

24、mcpy(temp,select0,strlen(select0);tempstrlen(select0)=0;for(i=1;i=count-1;i+) /*判斷輸入文法是否為L(zhǎng)L(1)文法*/ length=strlen(temp);if(lefti=lefti-1)merge(temp,selecti,1);if(strlen(temp)length+strlen(selecti)return(0);elsetemp0=0; memcpy(temp,selecti,strlen(selecti);tempstrlen(selecti)=0;return(1);/* 構(gòu)造分析表M*/void MM() int i,j,k,m;for(i=0;i=19;i+)for(j=0;j=19;j+)Mij=-1; i=strlen(termin); termini=#; /*將#加入終結(jié)符數(shù)組*/ termini+1=0;for(i=0;i=count-1;i+) for(m=0;m+)if(non_term=lefti)break; /*m為產(chǎn)生式左部非終結(jié)符的序號(hào)*/for(j=0;size_t(j)=0;n-) Sp+=rightmn; Sq+strlen(rightm)=0;