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LL(1) parser code

Dmytro Skorobahat'ko answered on October 23, 2023 Popularity 7/10 Helpfulness 2/10

answer LL(1) parser code

LL(1) parser code

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#include<iostream.h>
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>


int main()
     {
       char tmp[15];
       char str[15], csf[15]={"E"};
       int ssm=0, i, j, v, k, pos=0, a;
       clrscr();

       char pt[6][5][4]={"" , "i" , "+" , "*" , "/",
             "E", "TX", "n" , "n" , "n",
             "X", "n" , "TX", "n" , "" ,
             "T", "VY", "n" , "n" , "n",
             "Y", "n" , ""  , "VY", "" ,
             "V", "i" , "n" , "n" , "n"};
     cout<<"\n Enter An Expression: "<<endl;
     cin>>str;

     while(str[ssm]!='/')
    {
      pos=0;
      while(csf[pos]!='E' &&
        csf[pos]!='X' &&
        csf[pos]!='T' &&
        csf[pos]!='Y' &&
        csf[pos]!='V' && csf[pos]!='\0')
        pos++;
    if (csf[pos]=='\0')
       break;

    for(i=1; i<6; i++)
      {
        if(csf[pos]==pt[i][0][0])
           break;
      }
       for(j=1; j<5; j++)
      {
         if(str[ssm]==pt[0][j][0])
        break;
      }
       if(strcmp(pt[i][j], "n")==0)
       {
          cout<<"\n Null Value: "<<endl;
          cout<<pt[i][j];
          exit;
       }

       for(k=0; k<pos; k++)
       tmp[k]=csf[k];
       tmp[k]='\0';
       strcat(tmp, pt[i][j]);

       int l= strlen(tmp);
       k=l;

       for(a=pos+1; csf[a]!='\0'; a++, k++)
       tmp[k]=csf[a];
       strcpy(csf, tmp);
    }


       strcat(csf, "/");
     if(strcmp(csf, str)==0)
         {
         cout<<"\t csf"<<csf;

         cout<<"\tstr"<<str;

         cout<<"\n Valid Expression: "<<endl;


         }

     else
         {
          cout<<"\n Invalid Expression:"<<endl;

         }


      getch();
      return 0;
}

Popularity 7/10 Helpfulness 2/10 Language cpp

Source: cplusplus.com

Tags: c++

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Contributed on Oct 23 2023

Dmytro Skorobahat'ko

0 Answers Avg Quality 2/10

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