问题
I'm trying to understand how LR1 Parsers work but I came up with a strange problem: What if the grammar contains Epsilons? For instance: if I have the grammar:
S -> A
A -> a A | B
B -> a
It's clear how to start:
S -> .A
A -> .a A
A -> .B
... and so on
but I don't know how to do it for such a grammar:
S -> A
A -> a A a | \epsilon
Is it correct to do:
S -> .A
A -> .a A a
( A -> .\epsilon )
And then make this State in the DFA accepting?
Any help would really be appreciated!
回答1:
Yes, exactly (think of the epsilon as empty space, where there aren't two places for the dot at the sides).
In an LR(0) automaton, you would make the state accepting and reduce to A. However, due to the A->a A a production, there'd be a shift/reduce conflict.
In a LR(1) automaton, you would determine whether to shift or reduce using lookahead (a -> shift, anything in FOLLOW(A) -> reduce)
See the Wikipedia article
来源：https://stackoverflow.com/questions/486920/lr1-parser-and-epsilon