Re: converting this grammar to LL1

>
> E-> F = E
> E -> F
>
> F -> id| & id | ( E ) | E [ int ]
>
>
> then I apply another algorithm to eliminate left-recursions.
> then I apply another algorithm to left-factor this grammar.
>
>
> However, I use this tool LellRap to verify my resulted grammar. It is
> still ambiguous.
>


Well, I don't think your new grammar is ambiguous, but it doesn't mean
that it can be parsed by a LL1 automaton (which is probably what
LellRap
is trying to tell you). Let's try with two sentences:
"a=b" and "&a=b"
For "a=b", when I start the current token is id ("a") so I must derive
a rule that starts with F, but should I use E->F or E->F=E ? This is
where
the look-ahead token helps me: "=" tells me to use E->F=E.


For "&a=b", the current token is "&" and the look-ahead token is id,
which
does not allow to take decision between E->F and E->F=E. To take this
decision, I have to look more than just one token ahead (two in this
case).


I believe the following grammar is LL1 (basically I just eliminated
left recursions in your original grammar):
E -> E' F
F -> [ int ] F | = E' F | (epsilon)
E' -> id | & id | ( E )


Patrick.

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