how to find the first and follow values of grammar - grammar

Question

i have the following grammar and i would like to create the First & follow table. if i have a case that the first of non terminal is epsilon should i take also all the terminals that came after this non terminal from is rule?
S-> ABC
A->Aa/aB
B->Bb/epsilon
C->Cc/epsilon
and my question is:
in the first of C i need to get First(C) = {epsilon,c) and First(B) = {epsilon,b)?
i got the following results but still i think i have problems:
|first|follow
S |a |$
A |a |a
B |eps,b|b,a,$
C |eps,c|$

Answer 1

Finding the FIRST and Follow of a Grammar written in txt FILE:
#include<stdio.h>
#include<string.h>
#define size 10
int i,j,l,m,n=0,o,p,nv,z=0,x=0;
char str[size],temp,temp2[size],temp3[20],*ptr;
struct prod
{
char left_of_non_term[size],right_of_nonTerm[size][size],first[size],fol[size];
int n;
}pro[size];
int main()
{
FILE *f;
for(i=0;i<size;i++)
pro[i].n=0;
f=fopen("lab6.txt","r");
while(!feof(f))
{
fscanf(f,"%s",pro[n].left_of_non_term);
if(n>0)
{
if( strcmp(pro[n].left_of_non_term,pro[n-1].left_of_non_term) == 0 )
{
pro[n].left_of_non_term[0]='\0';
fscanf(f,"%s",pro[n-1].right_of_nonTerm[pro[n-1].n]);
pro[n-1].n++;
continue;
}
}
fscanf(f,"%s",pro[n].right_of_nonTerm[pro[n].n]);
pro[n].n++;
n++;
}
printf("\nGiven Grammar");
printf("\n-------------\n");
for(i=0;i<n;i++)
for(j=0;j<pro[i].n;j++)
printf("%s = %s\n",pro[i].left_of_non_term,pro[i].right_of_nonTerm[j]);
pro[0].first[0]='#';
for(i=0;i<n;i++)
{
for(j=0;j<pro[i].n;j++)
{
if( pro[i].right_of_nonTerm[j][0]<65 || pro[i].right_of_nonTerm[j][0]>90 )
{
pro[i].first[strlen(pro[i].first)]=pro[i].right_of_nonTerm[j][0];
}
else if( pro[i].right_of_nonTerm[j][0]>=65 && pro[i].right_of_nonTerm[j][0]<=90 )
{
temp=pro[i].right_of_nonTerm[j][0];
if(temp=='S')
pro[i].first[strlen(pro[i].first)]='#';
findter();
}
}
}
printf("-------------");
printf("\n\nFIRST\n");
for(i=0;i<n;i++)
{
printf("\n%s -> { ",pro[i].left_of_non_term);
for(j=0;j<strlen(pro[i].first);j++)
{
for(l=j-1;l>=0;l--)
if(pro[i].first[l]==pro[i].first[j])
break;
if(l==-1)
printf("%c ",pro[i].first[j]);
}
printf("}");
}
for(i=0;i<n;i++)
temp2[i]=pro[i].left_of_non_term[0];
pro[0].fol[0]='$';
for(i=0;i<n;i++)
{
for(l=0;l<n;l++)
{
for(j=0;j<pro[i].n;j++)
{
ptr=strchr(pro[l].right_of_nonTerm[j],temp2[i]);
if( ptr )
{
p=ptr-pro[l].right_of_nonTerm[j];
if(pro[l].right_of_nonTerm[j][p+1]>=65 && pro[l].right_of_nonTerm[j][p+1]<=90)
{
for(o=0;o<n;o++)
if(pro[o].left_of_non_term[0]==pro[l].right_of_nonTerm[j][p+1])
strcat(pro[i].fol,pro[o].first);
}
else if(pro[l].right_of_nonTerm[j][p+1]=='\0')
{
temp=pro[l].left_of_non_term[0];
if(pro[l].right_of_nonTerm[j][p]==temp)
continue;
if(temp=='S')
strcat(pro[i].fol,"$");
findfol();
}
else
pro[i].fol[strlen(pro[i].fol)]=pro[l].right_of_nonTerm[j][p+1];
}
}
}
}
printf("\n\n\n");
for(i=0;i<n;i++)
{
printf("\nFOLLOW (%s) -> { ",pro[i].left_of_non_term);
for(j=0;j<strlen(pro[i].fol);j++)
{
for(l=j-1;l>=0;l--)
if(pro[i].fol[l]==pro[i].fol[j])
break;
if(l==-1)
printf("%c",pro[i].fol[j]);
}
printf(" }");
}
printf("\n");
//getch();
}
void findter()
{
int k,t;
for(k=0;k<n;k++)
{
if(temp==pro[k].left_of_non_term[0])
{
for(t=0;t<pro[k].n;t++)
{
if( pro[k].right_of_nonTerm[t][0]<65 || pro[k].right_of_nonTerm[t][0]>90 )
pro[i].first[strlen(pro[i].first)]=pro[k].right_of_nonTerm[t][0];
else if( pro[k].right_of_nonTerm[t][0]>=65 && pro[k].right_of_nonTerm[t][0]<=90 )
{
temp=pro[k].right_of_nonTerm[t][0];
if(temp=='S')
pro[i].first[strlen(pro[i].first)]='#';
findter();
}
}
break;
}
}
}
void findfol()
{
int k,t,p1,o1,chk;
char *ptr1;
for(k=0;k<n;k++)
{
chk=0;
for(t=0;t<pro[k].n;t++)
{
ptr1=strchr(pro[k].right_of_nonTerm[t],temp);
if( ptr1 )
{
p1=ptr1-pro[k].right_of_nonTerm[t];
if(pro[k].right_of_nonTerm[t][p1+1]>=65 && pro[k].right_of_nonTerm[t][p1+1]<=90)
{
for(o1=0;o1<n;o1++)
if(pro[o1].left_of_non_term[0]==pro[k].right_of_nonTerm[t][p1+1])
{
strcat(pro[i].fol,pro[o1].first);
chk++;
}
}
else if(pro[k].right_of_nonTerm[t][p1+1]=='\0')
{
temp=pro[k].left_of_non_term[0];
if(pro[l].right_of_nonTerm[j][p]==temp)
continue;
if(temp=='S')
strcat(pro[i].fol,"$");
findfol();
chk++;
}
else
{
pro[i].fol[strlen(pro[i].fol)]=pro[k].right_of_nonTerm[t]
[p1+1];
chk++;
}
}
}
if(chk>0)
break;
}
}
Also Make a text File named it lab6.txt . and put grammers like below
S ABCDE
A a|0
B b|0
C c
D d|0
E e|0
Here space after NonTerminal Indicates the -> this sign and 0 indicates epsilon.