Compiler Design

Question

Confusion in follow of A,Please clarify.

Comments

$follow (A)=follow(S)$

$follow (S)=first (A)$

$first(A)=(a,\epsilon)$

replace the production $S\rightarrow bSA$ with $\epsilon$ we get $S\rightarrow bS, follow(S)=(\$,a)$

Answer 1

Non terminals	First	Follow
$S$	$\left \{ a,b \right \}$	$\left \{ \$,a \right \}$
$A$	$\left \{ a,\epsilon \right \}$	$\left \{\$ ,a \right \}$

 

Now the predictive parsing table :-

Non terminals	$a$	$b$	$\$$
$S$	$S\to aS$	$S\to bSA$
$A$	$A\to a$ $A\to \epsilon$		$A\to \epsilon$

So for $M[A,a]$ we have two entries 

$A\to a$

$A\to \epsilon$ .