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ASSIGNMENT- 6
1.Consider the following grammar rules: E->E+T|T, T->T*F|F, F->(E)|id where E, T and F are non-terminals and +, *, (, ), id are terminals (tokens). Design a shift-reduce parser for it. Solution: #include<bits/stdc++.h> using namespace std; struct grammer{ char p[20]; char prod[20]; }g[10]; int main() { cout<<"\t\t\t SHIFT REDUCE PARSER\t\t\t\n"; int i,stpos,j,k,l,m,o,p,f,r; int np,tspos,cr; cout<<"\nEnter Number of productions:"; cin>>np; char sc,ts[10]; cout<<"\nEnter productions:\n"; for(i=0;i<np;i++) {
cin>>ts; strncpy(g[i].p,ts,1); strcpy(g[i].prod,&ts[3]); } char ip[10]; cout<<"\nEnter Input:"; cin>>ip; int lip=strlen(ip); char stack[10]; stpos=0; i=0; //moving input sc=ip[i]; stack[stpos]=sc; i++;stpos++; cout<<"\n\nStack\t\tInput\t\tAction"; do { r=1; while(r!=0) { cout<<"\n"; for(p=0;p<stpos;p++) {
ts[tspos]=stack[l]; tspos++; } //now compare each possibility with production for(m=0;m<np;m++) { cr = strcmp(ts,g[m].prod); //if cr is zero then match is found if(cr==0) { for(l=k;l<10;l++) //removing matched part from stack { stack[l]='\0'; stpos--; } stpos=k; //concatinate the string strcat(stack,g[m].p); stpos++; r=2; } } } } //moving input
sc=ip[i]; stack[stpos]=sc; i++;stpos++; }while(strlen(stack)!=1 && stpos!=lip); if(strlen(stack)==1) { cout<<"\n\n \t\t\tSTRING IS ACCEPTED\t\t\t"; } else cout<<"\n\n \t\t\tSTRING IS REJECTED\t\t\t"; return 0; } Output:
/* ^ */ '>', '>','>','>','<','<','<','>','>',
/* i / '>', '>','>','>','>','e','e','>','>', / ( / '<', '<','<','<','<','<','<','>','e', / ) / '>', '>','>','>','>','e','e','>','>', / $ / '<', '<','<','<','<','<','<','<','>', }; int getindex(char c) { switch(c) { case '+':return 0; case '-':return 1; case '':return 2; case '/':return 3; case '^':return 4; case 'i':return 5; case '(':return 6; case ')':return 7; case '$':return 8; } }
int shift() { stack[++top]=*(input+i++); stack[top+1]='\0'; } int reduce() { int i,len,found,t; for(i=0;i<5;i++)//selecting handles { len=strlen(handles[i]); if(stack[top]==handles[i][0]&&top+1>=len) { found=1; for(t=0;t<len;t++) { if(stack[top-t]!=handles[i][t]) { found=0; break; } } if(found==1) { stack[top-t+1]='E'; top=top-t+1; strcpy(lasthandle,handles[i]);
input=(char)malloc(50sizeof(char)); printf("\nEnter the string\n"); scanf("%s",input); input=strcat(input,"$"); l=strlen(input); strcpy(stack,"$"); printf("\nSTACK\tINPUT\tACTION"); while(i<=l) { shift(); printf("\n"); dispstack(); printf("\t"); dispinput(); printf("\tShift"); if(prec[getindex(stack[top])][getindex(input[i])]=='>') { while(reduce()) { printf("\n"); dispstack(); printf("\t"); dispinput(); printf("\tReduced: E->%s",lasthandle); } } } if(strcmp(stack,"$E$")==0) printf("\nAccepted;");
else printf("\nNot Accepted;"); } Output:
beta = production[i][index+1]; printf("Grammar without left recursion:\n"); printf("%c->%c%c'", non_terminal, beta, non_terminal); printf("\n%c'->%c%c'|^\n", non_terminal, alpha, non_terminal); } else printf(" can't be reduced\n"); } else printf(" is not left recursive.\n"); index = 3; }return 0; } To eliminate left recursion: #include <bits/stdc++.h> using namespace std; struct production { char lf; char rt[10]; int prod_rear; int fl; }; struct production prodn[20],prodn_new[20]; //Creation of object //Variables Declaration for left factoring int b=-1,d,q,f,n,m=0,c=0; char terminal[20],nonterm[20],alpha[10],extra[10];
char epsilon='^'; void left_fact(vector &prod, int n) { int cnt, cnt3; char lp; string d = "->"; //Input of Productions cout<<"Note: Here ^ denotes epsillon\n"; for(cnt=0;cnt<=n-1;cnt++) { int pos = prod[cnt].find(d); lp = prod[cnt][0]; prodn[cnt].lf = lp; for(int i=pos+2; i<=prod[cnt].length()-1; i++) { prodn[cnt].rt[i-(pos+2)] = prod[cnt][i]; } prodn[cnt].prod_rear=strlen(prodn[cnt].rt); prodn[cnt].fl=0; } //Condition for left factoring int cnt1 = 0, cnt2 = 0; for(cnt1=0;cnt1<n;cnt1++) { for(cnt2=cnt1+1;cnt2<n;cnt2++) { if(prodn[cnt1].lf==prodn[cnt2].lf) { cnt=0;
cnt++; } if((prodn[cnt1].rt[cnt]==0)&&(m==0)) { int h=0; prodn_new[++b].lf=prodn[cnt1].lf; strcpy(prodn_new[b].rt,extra); //prodn_new[b].rt[p+1]=alpha[c]; prodn_new[b].rt[p+1]=prodn[cnt1].lf; //prodn_new[++b].lf=alpha[c]; prodn_new[++b].lf=prodn[cnt1].lf; prodn_new[b].rt[0]=epsilon; //prodn_new[++b].lf=alpha[c]; prodn_new[++b].lf=prodn[cnt1].lf; for(q=cnt;q<prodn[cnt2].prod_rear;q++) prodn_new[b].rt[h++]=prodn[cnt2].rt[q]; } if((prodn[cnt2].rt[cnt]==0)&&(m==0)) { int h=0; prodn_new[++b].lf=prodn[cnt1].lf; strcpy(prodn_new[b].rt,extra); //prodn_new[b].rt[p+1]=alpha[c]; prodn_new[b].rt[p+1]=prodn[cnt1].lf; //prodn_new[++b].lf=alpha[c]; prodn_new[++b].lf=prodn[cnt1].lf; prodn_new[b].rt[0]=epsilon; //prodn_new[++b].lf=alpha[c]; prodn_new[++b].lf=prodn[cnt1].lf; for(q=cnt;q<prodn[cnt1].prod_rear;q++)
prodn_new[b].rt[h++]=prodn[cnt1].rt[q]; } c++; m=0; } } } //Display of Output cout<<"\n After Left Factoring \n"; cout<<endl; for(cnt3=0;cnt3<=0;cnt3++) { cout<<"Production "<<cnt3+1<<" is: "; cout<<prodn_new[cnt3].lf; cout<<"->"; cout<<prodn_new[cnt3].rt<<"'"; cout<<endl<<endl; } for(cnt3=1;cnt3<=b;cnt3++) { cout<<"Production "<<cnt3+1<<" is: "; cout<<prodn_new[cnt3].lf<<"'"; cout<<"->"; cout<<prodn_new[cnt3].rt; cout<<endl<<endl; } cnt3 = b+2; for(int cnt4=0;cnt4<n;cnt4++) {
Output:
- Write a C program to compute FIRST and FOLLOW sets for a given grammar, and check whether the grammar is LL(1). Solution: #include "ctype.h" #include "string.h" #include "stdio.h" char gram[10][10]; char vFirst[10]; char nonT[10]; char vFollow[10]; int m = 0; int p; int i = 0; int j = 0; int elem[10]; int size; int fPt; int k = 0; int getGram() { char ch; int i; int j; int k; printf("\nEnter Number of Rule : "); scanf("%d", &size); printf("\nEnter Grammar as E=E+B \n"); for(i = 0; i < size; i++){ scanf("%s%c", gram[i], &ch); elem[i] = strlen(gram[i]); }
