Leading And Trailing Source Code in C Program Complier Design

First include the Necessary package. Declare the array and there function. Get the input by through for loop. I use the following function to perform the variable. int searchnt(char a) ,void push(char a),void installl(int a,int b), void installt(int a,int b).
Source Code Programming Leading And Trailing
#include<iostream.h>
#include<string.h>
#include<conio.h>
int nt,t,top=0;
char s[50],NT[10],T[10],st[50],l[10][10],tr[50][50];
int searchnt(char a)
{
int count=-1,i;
for(i=0;i<nt;i++)
{
if(NT[i]==a)
return i;
}
return count;
}
int searchter(char a)
{
int count=-1,i;
for(i=0;i<t;i++)
{
if(T[i]==a)
return i;
}
return count;
}
void push(char a)
{
s[top]=a;
top++;
}
char pop()
{
top--;
return s[top];
}
void installl(int a,int b)

{
if(l[a][b]=='f')
{
l[a][b]='t';
push(T[b]);
push(NT[a]);
}
}
void installt(int a,int b)
{
if(tr[a][b]=='f')
{
tr[a][b]='t';
push(T[b]);
push(NT[a]);
}
}

void main()
{
int i,s,k,j,n;
char pr[30][30],b,c;
clrscr();
cout<<"Enter the no of productions:";
cin>>n;
cout<<"Enter the productions one by one\n";
for(i=0;i<n;i++)
cin>>pr[i];
nt=0;
t=0;
for(i=0;i<n;i++)
{
if((searchnt(pr[i][0]))==-1)
NT[nt++]=pr[i][0];
}
for(i=0;i<n;i++)
{
for(j=3;j<strlen(pr[i]);j++)
{
if(searchnt(pr[i][j])==-1)
{
if(searchter(pr[i][j])==-1)
T[t++]=pr[i][j];
}
}
}
for(i=0;i<nt;i++)
{
for(j=0;j<t;j++)
l[i][j]='f';
}
for(i=0;i<nt;i++)
{
for(j=0;j<t;j++)

tr[i][j]='f';
}
for(i=0;i<nt;i++)
{
for(j=0;j<n;j++)
{
if(NT[(searchnt(pr[j][0]))]==NT[i])
{
if(searchter(pr[j][3])!=-1)
installl(searchnt(pr[j][0]),searchter(pr[j][3]));
else
{
for(k=3;k<strlen(pr[j]);k++)
{
if(searchnt(pr[j][k])==-1)
{
installl(searchnt(pr[j][0]),searchter(pr[j][k]));
break;
}
}
}
}
}
}
while(top!=0)
{
b=pop();
c=pop();
for(s=0;s<n;s++)
{
if(pr[s][3]==b)
installl(searchnt(pr[s][0]),searchter(c));
}
}
for(i=0;i<nt;i++)
{
cout<<"Leading["<<NT[i]<<"]"<<"\t{";
for(j=0;j<t;j++)
{
if(l[i][j]=='t')
cout<<T[j]<<",";
}
cout<<"}\n";
}


top=0;
for(i=0;i<nt;i++)
{
for(j=0;j<n;j++)
{
if(NT[searchnt(pr[j][0])]==NT[i])
{
if(searchter(pr[j][strlen(pr[j])-1])!=-1)
installt(searchnt(pr[j][0]),searchter(pr[j][strlen(pr[j])-1]));
else
{
for(k=(strlen(pr[j])-1);k>=3;k--)
{
if(searchnt(pr[j][k])==-1)
{
installt(searchnt(pr[j][0]),searchter(pr[j][k]));
break;
}
}
}
}
}
}
while(top!=0)
{
b=pop();
c=pop();
for(s=0;s<n;s++)
{
if(pr[s][3]==b)
installt(searchnt(pr[s][0]),searchter(c));
}
}
for(i=0;i<nt;i++)
{
cout<<"Trailing["<<NT[i]<<"]"<<"\t{";
for(j=0;j<t;j++)
{
if(tr[i][j]=='t')
cout<<T[j]<<",";
}
cout<<"}\n";
}
getch();
}
Output Leading And Trailing Source Code in C Programming
Enter the no of productions:6
Enter the productions one by one
E->E+E
E->T
T->T*F
F->(E)
T->F
F->i
Leading[E] {+,*,(,i,}
Leading[T] {*,(,i,}
Leading[F] {(,i,}
Trailing[E] {+,*,),i,}
Trailing[T] {*,),i,}
Trailing[F] {),i,}

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