LL(1) 手工构造预测语法分析程序

#include <iostream> #include <stack> using namespace std; char msg[50]; class mystack{ public: char element[100]; int size; void clear() { size=0; } int length() { return size; } char top() { if(size>0) return element[size-1]; } void pop() { if(size>0) { size--; element[size]='\0'; } } void push(char st) { element[size++]=st; element[size]='\0'; } mystack() { size=0; } }; mystack sc; // void printstack() { printf("%s\t",sc.element); } // char s[80];//输入的字符串 int l;//输入字符串的长度 struct any { char a1[10]; char a2[10]; char a3[10]; char a4[10]; char a5[10]; char a6[10]; }; struct any S={"E","E","TA","TA","E","E"}; struct any A={"E","E","UB","UB","E","E"}; struct any B={"E","E","m",")s(","E","E"}; struct any U={"","UB*","TA","TA","E",""}; struct any T={"TA+","E","E","E","E",""}; //////////////////////////////////////////////////////// bool comp(int& b) { char d; //int i; d=sc.top(); if(d=='+'||d=='*'||d=='m'||d=='('||d==')') { if(s[b]==d) { msg[0]=s[b]; strcpy(msg+1,"匹配"); b++; sc.pop(); return true; } else { printf("ERROR!"); return false; } } //---------------------------------------------- else if(d=='S') { if(s[b]=='m') { strcpy(msg,"S->AT"); sc.pop(); for(int i=0;i<strlen(S.a3);i++) { sc.push(S.a3[i]); } //strcpy(msg,"S->AT"); return true; } else if(s[b]=='(') { strcpy(msg,"S->AT"); sc.pop(); for(int i=0;i<strlen(S.a4);i++) { sc.push(S.a4[i]); } //strcpy(msg,"S->AT"); return true; } else { printf("ERROR!"); return false; } } //--------------------------------------------- else if(d=='A') { if(s[b]=='m') { strcpy(msg,"A->BU"); sc.pop(); for(int i=0;i<strlen(A.a3);i++) { sc.push(A.a3[i]); } //strcpy(msg,"A->BU"); return true; } else if(s[b]=='(') { strcpy(msg,"A->BU"); sc.pop(); for(int i=0;i<strlen(A.a4);i++) { sc.push(A.a4[i]); } //strcpy(msg,"A->BU"); return true; } else { printf("ERROR!"); return false; } } //------------------------------------------- else if(d=='B') { if(s[b]=='m') { sc.pop(); for(int i=0;i<strlen(B.a3);i++) { sc.push(B.a3[i]); } strcpy(msg,"B->m"); return true; } else if(s[b]=='(') { sc.pop(); for(int i=0;i<strlen(B.a4);i++) { sc.push(B.a4[i]); } strcpy(msg,"B->(S)"); return true; } else { printf("ERROR!"); return false; } } //---------------------------------------------- else if(d=='U') { if(s[b]=='+') { sc.pop(); strcpy(msg,"S->$"); return true; } else if(s[b]=='*') { sc.pop(); for(int i=0;i<strlen(U.a2);i++) { sc.push(U.a2[i]); } strcpy(msg,"S->*BU"); return true; } else if(s[b]=='#') { sc.pop(); strcpy(msg,"U->$"); return true; } else { printf("ERROR!"); return false; } } //------------------------------------------------- else if(d=='T') { if(s[b]=='+') { sc.pop(); for(int i=0;i<strlen(T.a1);i++) { sc.push(T.a1[i]); } strcpy(msg,"T->+AT"); return true; } else if(s[b]=='#') { sc.pop(); strcpy(msg,"T->$"); return true; } else { printf("ERROR!"); return false; } } //--------------------------------------------------- else if(d=='#') { printf("接受"); return false; } //-------------------------------------------------- else { printf("ERROR!"); return false; } } /////////////////////////////////////////////// void main() { while(true) { sc.clear(); printf("please input string(exit to quit):"); gets(s); if(strcmp(s,"exit")==0) break; l=strlen(s);//取得字符串的长度 s[l]='#';//对输入的字符串最后补充一个#号 l++; s[l]='\0'; sc.push('#'); sc.push('S'); int b; b=0; printf("用预测分析法分析符号串的过程："); printf("\n"); printf("step\tStack\tString\tRule\n"); int st=1; printf("%d\t",st); printstack();//输出栈内元素 printf("%s\t",&s[b]);//输出剩余字符串的内容 //printf("S->AT\n"); while(b<l&&comp(b)) { printf("%s\n",msg); st++; printf("%d\t",st); printstack();//输出栈内元素 printf("%s\t",&s[b]);//输出剩余字符串的内容 } printf("\n"); } }