Conversion of a Regular Expression to NFA Algorithm Source code C programmingCS342 Compiler Lab Source code Algorithm C Programming

#include

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

{

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();

}

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