Conversion of a Regular Expression to NFA Algorithm Source code C programming

CS342 Compiler Lab Source code Algorithm C Programming
#include<stdio.h>
#include<conio.h>
void main()
{
char reg[20];
int q[20][3],i,j,len,a,b;
clrscr();
for(a=0;a<20;a++)
{
for(b=0;b<3;b++)
{
q[a][b]=0;
}
}
printf("Regular expression: \n");
scanf("%s",reg);
len=strlen(reg);
i=0;
j=1;
while(i<len)
{
if(reg[i]=='a'&®[i+1]!='/'&®[i+1]!='*')
{
q[j][0]=j+1;
j++;
}
if(reg[i]=='b'&®[i+1]!='/'&®[i+1]!='*')
{
q[j][1]=j+1;
j++;
}
if(reg[i]=='e'&®[i+1]!='/'&®[i+1]!='*')
{
q[j][2]=j+1;
j++;
}
if(reg[i]=='a'&®[i+1]=='/'&®[i+2]=='b')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][0]=j+1;
j++;
q[j][2]=j+3;
j++;
q[j][1]=j+1;
j++;
q[j][2]=j+1;
j++;
i=i+2;
}
if(reg[i]=='b'&®[i+1]=='/'&®[i+2]=='a')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][1]=j+1;
j++;
q[j][2]=j+3;
j++;
q[j][0]=j+1;
j++;
q[j][2]=j+1;
j++;
i=i+2;
}
if(reg[i]=='a'&®[i+1]=='*')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][0]=j+1;
j++;
q[j][2]=((j+1)*10)+(j-1);
j++;
}
if(reg[i]=='b'&®[i+1]=='*')
{
q[j][2]=((j+1)*10)+(j+3);
j++;
q[j][1]=j+1;
j++;
q[j][2]=((j+1)*10)+(j-1);
j++;
}
if(reg[i]==')'&®[i+1]=='*')
{
q[0][2]=((j+1)*10)+1;
q[j][2]=((j+1)*10)+1;
j++;
}
i++;
}
printf("Transition function \n");
for(i=0;i<=j;i++)
{
if(q[i][0]!=0)
printf("\n q[%d,a]-->%d",i,q[i][0]);
if(q[i][1]!=0)
printf("\n q[%d,b]-->%d",i,q[i][1]);
if(q[i][2]!=0)
{
if(q[i][2]<10)
printf("\n q[%d,e]-->%d",i,q[i][2]);
else
printf("\n q[%d,e]-->%d & %d",i,q[i][2]/10,q[i][2]%10);
}
}
getch();
}
This is Conversion of a Regular Expression to NFA Algorithm

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