编译原理实验代码.doc

实验任务 完成以下正则文法所描述的Pascal语言子集单词符号的词法分析程序。字母 字母 数字 数字 数字 + * ; () =* = : /该语言的保留字 ：begin end if then else for do while and or not 说明： 1 该语言大小写不敏感。 2 字母为a-z A-Z，数字为0-9。3可以对上述文法进行扩充和改造。4 /*/为程序的注释部分。设计要求1、给出各单词符号的类别编码。2、词法分析程序应能发现输入串中的错误。3、词法分析作为单独一遍编写，词法分析结果为二元式序列组成的中间文件。4、设计两个测试用例（尽可能完备），并给出测试结果。 demo.cpp #include #include #include #include demo.hchar token20;int lookup(char *token) for (int i = 0; i 11; i+) if (strcmp(token, KEY_WORDSi) = 0) return i+1; return 0;char getletter(FILE *fp) return tolower(fgetc(fp);void out(FILE *fp, int c, char *value) fprintf(fp, %d,%sn, c, value);void report_error(FILE *fp, char ch) fprintf(fp, There must be some error.n); fprintf(fp, %c.n, ch); fprintf(fp, n);void scanner(FILE *infile, FILE *outfile) char ch; int i, c; do do ch = getletter(infile); while(isspace(ch); if (isalpha(ch) token0 = ch; ch = getletter(infile); i = 1; while (isalnum(ch) tokeni = ch; i+; ch = getletter(infile); tokeni = 0; fseek(infile, -1, 1); c = lookup(token); if (c = 0) out(outfile, ID, token); else out(outfile, c, ); else if (isdigit(ch) token0 = ch; ch = getletter(infile); i = 1; while (isdigit(ch) tokeni = ch; i+; ch = getletter(infile); tokeni = 0; fseek(infile, -1, 1); out(outfile, INT, token); else switch(ch) case ) out(outfile, NE, ); else fseek(infile, -1, 1); out(outfile, LT, ); break; case =: out(outfile, EQ, ); break; case : ch = getletter(infile); if (ch = =) out(outfile, GE, ); else fseek(infile, -1, 1); out(outfile, GT, ); break; case : ch = getletter(infile); if (ch = =) out(outfile, FU, ); else fseek(infile, -1, 1); out(outfile, MAO, ); break; case /: ch = getletter(infile); if (ch = *) out(outfile, ZHU, ); else fseek(infile, -1, 1); out(outfile, XIE, ); break; case +: out(outfile, JIA, ); break; case -: out(outfile, JIAN, ); break; case *: out(outfile, CHEN, ); break; case ;: out(outfile, FEN, ); break; case (: out(outfile, ZUO, ); break; case ): out(outfile, YOU, ); break; default: if (ch != EOF) report_error(outfile, ch); break; while(ch != EOF); return;void main() FILE *in, *out; in = fopen(pascal.txt, r); out = fopen(result.txt, w); scanner(in, out); fprintf(out, 0,#); fclose(in); fclose(out); printf(Finished!n); 本文来自CSDN博客，转载请标明出处：/ffee/archive/2006/05/31/766483.aspx实验任务1、实现LL(1)分析中控制程序（表驱动程序）；2、完成以下描述算术表达式的LL(1)文法的LL(1)分析程序（LL(1)分析表见教材）。GE: ETE EATE| TFT TMFT| F (E)|i A+|- M*|/ 说明：终结符号i为用户定义的简单变量,即标识符的定义。设计要求1、输入串应是词法分析的输出二元式序列，即某算术表达式“实验项目一”的输出结果。输出为输入串是否为该文法定义的算术表达式的判断结果。2、LL(1)分析过程应能发现输入串出错。3、设计两个测试用例（尽可能完备，正确和出错），并给出测试结果。demo.cpp #include #include #include #include demo.h#define SIZE 100char stackSIZE;int bottom = 0, top = 0;void push(char ch) if (top != SIZE) stacktop = ch; top+; char pop() if (top != bottom) top-; stacktop = 0; return stacktop; return -2;int get_id(FILE *fp) char temp100; int id; fscanf(fp, %d, &id); fgets(temp, 100, fp); return id;void translate(int id, char *a) switch(id) case 0: *a = #; break; case 12: *a = i; break; case 22: *a = +; break; case 23: *a = -; break; case 24: *a = *; break; case 21: *a = /; break; bool is_terminate(char ch) if (ch = i | ch = + | ch = - | ch = * | ch = / | ch = #) return true; return false;void trans(char ch, char a, int *i, int *j) switch (ch) case E: *i = 0; break; case e: *i = 1; break; case T: *i = 2; break; case t: *i = 3; break; case F: *i = 4; break; case A: *i = 5; break; case M: *i = 6; break; switch (a) case i: *j = 0; break; case +: *j = 1; break; case -: *j = 2; break; case *: *j = 3; break; case /: *j = 4; break; case (: *j = 5; break; case ): *j = 6; break; case #: *j = 7; break; void search_table(char ch, char a) int i, j; trans(ch, a, &i, &j); char temp20 = 0; strcpy(temp, TABLEij); if (strcmp(temp, ) = 0) pop(); return; else if (strcmp(temp, ) != 0) pop(); for (unsigned ii = 0; ii strlen(temp); ii+) push(tempii); return; else printf(Error!n); exit(0); void main() FILE *in; int id = 0; char a = 0; in = fopen(result.txt, r); push(#); push(E); id = get_id(in); translate(id, &a); while (true) if (is_terminate(stacktop-1) if (stacktop-1 = a & a = #) printf(success.n); return; else if (stacktop-1 = a) pop(); id = get_id(in); translate(id, &a); else printf(Error!n); return; else if (!is_terminate(stacktop-1) search_table(stacktop-1, a); fclose(in); 本文来自CSDN博客，转载请标明出处：/ffee/archive/2006/05/31/766487.aspx实验任务 完成以下描述算术表达式的LL(1)文法的递归下降分析程序GE: ETE EATE| TFT TMFT| F (E)|i A+|- M*|/ 说明：终结符号i为用户定义的简单变量,即标识符的定义。设计要求1、 输入串应是词法分析的输出二元式序列，即某算术表达式“实验项目一”的输出结果。输出为输入串是否为该文法定义的算术表达式的判断结果。2、 递归下降分析程序应能发现输入串出错。3、 设计两个测试用例（尽可能完备，正确和出错），并给出测试结果。demo.cpp bool E() if (T() if (E_prime() return true; return false;bool E_prime() if (A() if (T() if (E_prime() return true; else return false; else return false; return true;bool T() if (F() if (T_prime() return true; return false;bool T_prime() if (M() if (F() if (T_prime() return true; else return false; else return false; return true;bool F() if (a = () advance(); if (E() if (a = ) advance(); return true; else if (a = i) advance(); return true; return false;bool A() if (a = +) advance(); return true; else if (a = -) advance(); return true; return false;bool M() if (a = *) advance(); return true; else if (a = /) advance(); return true; return false;本文来自CSDN博客，转载请标明出处：/ffee/archive/2006/05/31/766742.aspx编译原理实验源代码 词法分析，LL（1），逆波兰式，LR(1) 1、词法分析/*cifa fenxi chengxu*/#include #include #include #include #include #define NULL 0FILE *fp;char cbuffer;char *key8=if,else,for,while,do,return,break,continue;char *border6=,;,(,);char *arithmetic4=+,-,*,/;char *relation6=,=,;char *consts20;char *label20;int constnum=0,labelnum=0;int search(char searchchar,int wordtype) int i=0; switch (wordtype) case 1:for (i=0;i=7;i+) if (strcmp(keyi,searchchar)=0) return(i+1); case 2:for (i=0;i=5;i+) if (strcmp(borderi,searchchar)=0) return(i+1); return(0); case 3:for (i=0;i=3;i+) if (strcmp(arithmetici,searchchar)=0) return(i+1); return(0); case 4:for (i=0;i=5;i+) if (strcmp(relationi,searchchar)=0) return(i+1); return(0); case 5:for (i=0;i=constnum;i+) if (strcmp(constsi,searchchar)=0) return(i+1); constsi-1=(char *)malloc(sizeof(searchchar); strcpy(constsi-1,searchchar); constnum+; return(i); case 6:for (i=0;i=labelnum;i+) if (strcmp(labeli,searchchar)=0) return(i+1); labeli-1=(char *)malloc(sizeof(searchchar); strcpy(labeli-1,searchchar); labelnum+; return(i); char alphaprocess(char buffer) int atype; int i=-1; char alphatp20; while (isalpha(buffer)|(isdigit(buffer) alphatp+i=buffer; buffer=fgetc(fp); alphatpi+1=0; if (atype=search(alphatp,1) printf(%s (1,%d)n,alphatp,atype-1); else atype=search(alphatp,6); printf(%s (6,%d)n,alphatp,atype-1); return(buffer);char digitprocess(char buffer) int i=-1; char digittp20; int dtype; while (isdigit(buffer) digittp+i=buffer; buffer=fgetc(fp); digittpi+1=0; dtype=search(digittp,5); printf(%s (5,%d)n,digittp,dtype-1); return(buffer);char otherprocess(char buffer) int i=-1; char othertp20; int otype,otypetp; othertp0=buffer; othertp1=0; if (otype=search(othertp,3) printf(%s (3,%d)n,othertp,otype-1); buffer=fgetc(fp); goto out; if (otype=search(othertp,4) buffer=fgetc(fp); othertp1=buffer; othertp2=0; if (otypetp=search(othertp,4) printf(%s (4,%d)n,othertp,otypetp-1); goto out; else othertp1=0; printf(%s (4,%d)n,othertp,otype-1); goto out; if (buffer=:) buffer=fgetc(fp); if (buffer=) printf(:= (2,2)n); buffer=fgetc(fp); goto out; else if (otype=search(othertp,2) printf(%s (2,%d)n,othertp,otype-1); buffer=fgetc(fp); goto out; if (buffer!=n)&(buffer!= ) printf(%c error,not a wordn,buffer); buffer=fgetc(fp);out: return(buffer);void main() int i; for (i=0;i=20;i+) labeli=NULL; constsi=NULL; ; if (fp=fopen(example.c,r)=NULL) printf(error); else cbuffer = fgetc(fp); while (cbuffer!=EOF) if (isalpha(cbuffer) cbuffer=alphaprocess(cbuffer); else if (isdigit(cbuffer) cbuffer=digitprocess(cbuffer); else cbuffer=otherprocess(cbuffer); printf(overn); getchar(); 2、LL（1）/*LL(1)分析法源程序，只能在VC+中运行*/#include#include#include#includechar A20;/*分析栈*/char B20;/*剩余串*/char v120=i,+,*,(,),#;/*终结符*/char v220=E,G,T,S,F;/*非终结符 */int j=0,b=0,top=0,l;/*L为输入串长度 */typedef struct type/*产生式类型定义 */ char origin;/*大写字符*/ char array5;/*产生式右边字符 */ int length;/*字符个数 */type;type e,t,g,g1,s,s1,f,f1;/*结构体变量*/type C1010;/*预测分析表 */void print()/*输出分析栈*/ int a;/*指针*/ for(a=0;a=top+1;a+) printf(%c,Aa); printf(tt);/*print*/void print1()/*输出剩余串*/ int j; for(j=0;jb;j+)/*输出对齐符*/ printf( ); for(j=b;j=l;j+) printf(%c,Bj); printf(ttt);/*print1*/void main() int m,n,k=0,flag=0,finish=0; char ch,x; type cha;/*用来接受Cmn*/ /*把文法产生式赋值结构体*/ e.origin=E; strcpy(e.array,TG); e.length=2; t.origin=T; strcpy(t.array,FS); t.length=2; g.origin=G; strcpy(g.array,+TG); g.length=3; g1.origin=G; g1.array0=; g1.length=1; s.origin=S; strcpy(s.array,*FS); s.length=3; s1.origin=S; s1.array0=; s1.length=1; f.origin=F; strcpy(f.array,(E); f.length=3; f1.origin=F; f1.array0=i; f1.length=1; for(m=0;m=4;m+)/*初始化分析表*/ for(n=0;n=5;n+) Cmn.origin=N;/*全部赋为空*/ /*填充分析表*/ C00=e;C03=e; C11=g;C14=g1;C15=g1; C20=t;C23=t; C31=s1;C32=s;C34=C35=s1; C40=f1;C43=f; printf(提示:本程序只能对由i,+,*,(,)构成的以#结束的字符串进行分析,n); printf(请输入要分析的字符串:); do/*读入分析串*/ scanf(%c,&ch); if (ch!=i) &(ch!=+) &(ch!=*)&(ch!=()&(ch!=)&(ch!=#) printf(输入串中有非法字符n); exit(1); Bj=ch; j+; while(ch!=#); l=j;/*分析串长度*/ ch=B0;/*当前分析字符*/ Atop=#; A+top=E;/*#,E进栈*/ printf(步骤tt分析栈 tt剩余字符 tt所用产生式 n); do x=Atop-;/*x为当前栈顶字符*/ printf(%d,k+); printf(tt); for(j=0;j=5;j+)/*判断是否为终结符*/ if(x=v1j) flag=1; break; if(flag=1)/*如果是终结符*/ if(x=#) finish=1;/*结束标记*