编译原理实验二语法分析器LL(1)实现.doc

【精品文档】如有侵权，请联系网站删除，仅供学习与交流编译原理实验二语法分析器LL(1)实现.精品文档.编译原理程序设计实验报告表达式语法分析器的设计班级：计算机1306班姓名：张涛学号：20133967实验目标：用LL（1）分析法设计实现表达式语法分析器实验内容：概要设计：通过对实验一的此法分析器的程序稍加改造，使其能够输出正确的表达式的token序列。然后利用LL（1）分析法实现语法分析。数据结构：int op=0; /当前判断进度char ch; /当前字符char nowword10="" /当前单词char operate4='+','-','*','/' /运算符char bound2='(',')' /界符struct Token int code; char ch10; /Token定义struct Token tokenlist50; /Token数组struct Token tokentemp; /临时Token变量struct Stack /分析栈定义 char *base; char *top; int stacksize;分析表及流程图关键函数：int IsLetter(char ch) /判断ch是否为字母int IsDigit(char ch) /判断ch是否为数字int Iskey(char *string) /判断是否为关键字int Isbound(char ch) /判断是否为界符int Isboundnum(char ch) /给出界符所在token值int init(STack *s) /栈初始化int pop(STack *s,char *ch) /弹栈操作int push(STack *s,char ch) /压栈操作void LL1(); /分析函数源程序代码：（加入注释）#include<stdio.h>#include<string.h>#include<ctype.h>#include<windows.h>#include <stdlib.h>int op=0; /当前判断进度char ch; /当前字符char nowword10="" /当前单词char operate4='+','-','*','/' /运算符char bound2='(',')' /界符struct Token int code; char ch10; /Token定义struct Token tokenlist50; /Token数组struct Token tokentemp; /临时Token变量struct Stack /分析栈定义 char *base; char *top; int stacksize;typedef struct Stack STack;int init(STack *s) /栈初始化 (*s).base=(char*)malloc(100*sizeof(char); if(!(*s).base) exit(0); (*s).top=(*s).base; (*s).stacksize=100; printf("初始化栈n"); return 0;int pop(STack *s,char *ch) /弹栈操作 if(*s).top=(*s).base) printf("弹栈失败n"); return 0; (*s).top-; *ch=*(*s).top); printf("%c",*ch); return 1;int push(STack *s,char ch) /压栈操作 if(*s).top-(*s).base>=(*s).stacksize) (*s).base=(char*)realloc(*s).base,(*s).stacksize+10)*sizeof(char); if(!(*s).base) exit(0); (*s).top=(*s).base+(*s).stacksize; (*s).stacksize+=10; *(*s).top=ch; *(*s).top+; return 1;void LL1();int IsLetter(char ch) /判断ch是否为字母 int i;for(i=0;i<=45;i+)if (ch>='a'&&ch<='z')|(ch>='A'&&ch<='Z')return 1; return 0;int IsDigit(char ch) /判断ch是否为数字 int i; for(i=0;i<=10;i+)if (ch>='0'&&ch<='9') return 1; return 0;int Isbound(char ch) /判断是否为界符 int i; for(i=0;i<2;i+) if(ch=boundi) return i+1; return 0;int Isoperate(char ch) /判断是否为运算符 int i; for(i=0;i<4;i+) if(ch=operatei) return i+3; return 0;int main()FILE *fp;int q=0,m=0;char sour200=" "printf("请将源文件置于以下位置并按以下方式命名：F:2.txtn");if(fp=fopen("F:2.txt","r")=NULL) printf("文件未找到！n"); else while(!feof(fp) if(isspace(ch=fgetc(fp); else sourq=ch; q+; int p=0; printf("输入句子为：n"); for(p;p<=q;p+) printf("%c",sourp); printf("n"); int state=0,nowlen=0; BOOLEAN OK=TRUE,ERR=FALSE; int i,flagpoint=0; for(i=0;i<q;i+) if(souri='#') tokenlistm.code='#' switch(state) case 0: ch=souri; if(Isbound(ch) if(ERR) printf("无法识别n"); ERR=FALSE; OK=TRUE; else if(!OK) printf("<10,%s>标识符n",nowword); tokentemp.code=10; tokentemp.ch10=nowword10; tokenlistm=tokentemp; m+; OK=TRUE; state=4; else if(IsDigit(ch) if(OK) memset(nowword,0,strlen(nowword); nowlen=0; nowwordnowlen=ch; nowlen+; state=3; OK=FALSE; break; else nowwordnowlen=ch; nowlen+; else if(IsLetter(ch) if(OK) memset(nowword,0,strlen(nowword); nowlen=0; nowwordnowlen=ch; nowlen+; OK=FALSE; else nowwordnowlen=ch; nowlen+; else if(Isoperate(ch) if(!OK) printf("<10,%s>标识符n",nowword); tokentemp.code=10; tokentemp.ch10=nowword10; tokenlistm=tokentemp; m+; OK=TRUE; printf("<%d,%c>运算符n",Isoperate(ch),ch); tokentemp.code=Isoperate(ch); tokentemp.ch10=ch; tokenlistm=tokentemp; m+; break; case 3: if(IsLetter(ch) printf("错误n"); nowwordnowlen=ch; nowlen+; ERR=FALSE; state=0; break; if(IsDigit(ch=souri) nowwordnowlen=ch; nowlen+; else if(souri='.'&&flagpoint=0) flagpoint=1; nowwordnowlen=ch; nowlen+; else printf("<20,%s>数字n",nowword); i-; state=0; OK=TRUE; tokentemp.code=20; tokentemp.ch10=nowword10; tokenlistm=tokentemp; m+; break; case 4: i-; printf("<%d,%c>界符n",Isbound(ch),ch); tokentemp.code=Isbound(ch); tokentemp.ch10=ch; tokenlistm=tokentemp; m+; state=0; OK=TRUE; break; printf("tokenlist值为%dn",m); int t=0; tokenlistm+1.code='r' m+; for(t;t<m;t+) printf("tokenlist%d值为%dn",t,tokenlistt.code); LL1(); printf("tokenlist值为%dn",m); if(op+1=m) printf("OK!"); else printf("WRONG!"); return 0;void LL1() STack s; init(&s); push(&s,'#'); push(&s,'E'); char ch; int flag=1; do pop(&s,&ch); printf("输出栈顶为%cn",ch); printf("输出栈顶为%dn",ch); printf("当前p值为%dn",op); if(ch='(')|(ch=')')|(ch='+')|(ch='-')|(ch='*')|(ch='/')|(ch=10)|(ch=20) if(tokenlistop.code=1|tokenlistop.code=20|tokenlistop.code=10|tokenlistop.code=2|tokenlistop.code=3|tokenlistop.code=4|tokenlistop.code=5|tokenlistop.code=6) op+; else printf("WRONG!"); exit(0); else if(ch='#') if(tokenlistop.code=0) flag=0; else printf("WRONG!"); exit(0); else if(ch='E') printf("进入En"); if(tokenlistop.code=10|tokenlistop.code=20|tokenlistop.code=1) push(&s,'R'); printf("将R压入栈n"); push(&s,'T'); else if(ch='R') printf("进入Rn"); if(tokenlistop.code=3|tokenlistop.code=4) push(&s,'R'); push(&s,'T'); printf("将T压入栈n"); push(&s,'+'); if(tokenlistop.code=2|tokenlistop.code=0) else if(ch='T') printf("进入Tn"); if(tokenlistop.code=10|tokenlistop.code=20|tokenlistop.code=1) push(&s,'Y'); push(&s,'F'); else if(ch='Y') printf("进入Yn"); if(tokenlistop.code=5|tokenlistop.code=6) push(&s,'Y'); push(&s,'F'); push(&s,'*'); else if(tokenlistop.code=3|tokenlistop.code=2|tokenlistop.code=0|tokenlistop.code=4) else if(ch='F') printf("进入Fn"); if(tokenlistop.code=10|tokenlistop.code=20) push(&s,10); if(tokenlistop.code=1) push(&s,')'); push(&s,'E'); push(&s,'('); else printf("WRONG!"); exit(0); while(flag);程序运行结果：（截屏）输入：(Aa+Bb)*(88.2/3)#注：如需运行请将文件放置F盘，并命名为：2.txt输出：思考问题回答： LL（1）分析法的主要问题就是要正确的将文法化为LL（1）文法。在程序实现时将分析表的动作逐一实现即可。