First and follow

[morshruti]

hey i am a student and i want a C program to compute the first and follow of any given productions.Plzzz rply soon......

[weaknessforcats]

Please read the posting guidelines.

I can help with specific programming and design issues but I cannot provide complete solutions or act as a tutor.

[JosAH]

Since you're repeating your previous question, I repeat my question to you:
do you know how to compute the first and follow sets given a grammar?

kind regards,

Jos

[curiously enough]

"i want a C program to compute the first and follow of any given productions."
say what?

[donbock]

"i want a C program to compute the first and follow of any given productions."
say what?

Finding the first and follow sets of a given grammar are things you would have to do if you were writing your own lex or flex.

[JosAH]

Finding the first and follow sets of a given grammar are things you would have to do if you were writing your own lex or flex.

Make that 'yacc' or 'bison' etc. Lex and flex use a simple NFA ---> DFA construct
(Non - Deterministic Finite Automaton) to build their lexical analyzer tables.

kind regards,

Jos

[nirjhor003]

#include<stdio.h>
#include<ctype.h>
char a[8][8];
int n=5 ;
struct firTab
{
int n;
char firT[5];
};
struct folTab
{
int n;
char folT[5];
};
struct folTab follow[5];
struct firTab first[5];
int col;
void findFirst(char,char);
void findFollow(char,char);
void folTabOperation(char,char);
void firTabOperation(char,char);
void main()
{
int i,j,c=0,cnt=0;
char ip;
char b[8];
printf("\n\t\t======FIRST AND FOLLOW SET====== \n\nEnter Production Amount : ");
scanf("%d",&n);
printf("\n\tenter %d productions in format E->T+G\n",n);
printf("\t------------------------------------\n");
for(i=0; i<n; i++)
{
scanf("%s",&a[i]);
}
for(i=0; i<n; i++)
{
c=0;
for(j=0; j<i+1; j++)
{
if(a[i][0] == b[j])
{
c=1;
break;
}
}
if(c !=1)
{
b[cnt] = a[i][0];
cnt++;
}

}
printf("\n");

for(i=0; i<cnt; i++)
{
col=1;
first[i].firT[0] = b[i];
first[i].n=0;
findFirst(b[i],i);
}
for(i=0; i<cnt; i++)
{
col=1;
follow[i].folT[0] = b[i];
follow[i].n=0;
findFollow(b[i],i);
}

printf("\n");
for(i=0; i<cnt; i++)
{
for(j=0; j<=first[i].n; j++)
{
if(j==0)
{
printf("First(%c) : {",first[i].firT[j]);
}
else
{
printf(" %c",first[i].firT[j]);
}
}
printf(" } ");
printf("\n");
}
printf("\n");
for(i=0; i<cnt; i++)
{
for(j=0; j<=follow[i].n; j++)
{
if(j==0)
{
printf("Follow(%c) : {",follow[i].folT[j]);
}
else
{
printf(" %c",follow[i].folT[j]);
}
}
printf(" } ");

printf("\n");
}

}
void findFirst(char ip,char pos)
{
int i;
for(i=0; i<n; i++)
{
if(ip == a[i][0])
{
if(isupper(a[i][3]))
{
findFirst(a[i][3],pos);
}
else
{

first[pos].firT[col]=a[i][3];
first[pos].n++;
col++;
}
}
}
}
void findFollow(char ip,char row)
{
int i,j;
if(row==0 && col==1)
{
follow[row].folT[col]= '$';
col++;
follow[row].n++;
}
for(i=0; i<n; i++)
{
for(j=3; j<7; j++)
{
if(a[i][j] == ip)
{
if(a[i][j+1] == '\0')
{
if(a[i][j] != a[i][0])
{
folTabOperation(a[i][0],row);
}
}
else if(isupper(a[i][j+1]))
{
if(a[i][j+1] != a[i][0])
{
firTabOperation(a[i][j+1],row);

}
}
else
{
follow[row].folT[col] = a[i][j+1];
col++;
follow[row].n++;


}
}
}
}
}
void folTabOperation(char ip,char row)
{
int i,j;
for(i=0; i<5; i++)
{
if(ip == follow[i].folT[0])
{
for(j=1; j<=follow[i].n; j++)
{
follow[row].folT[col] = follow[i].folT[j];
col++;
follow[row].n++;
}
}
}
}
void firTabOperation(char ip,char row)
{
int i,j;
for(i=0; i<5; i++)
{
if(ip == first[i].firT[0])
{
for(j=1; j<=first[i].n; j++)
{
if(first[i].firT[j] != '0')
{
follow[row].folT[col] = first[i].firT[j];
follow[row].n++;
col++;
}
else
{
folTabOperation(ip,row);
}
}
}
}

}