Re: eliminating left-recursion

Chris Dodd <cdodd@acm.org>8 Jan 2006 11:38:39 -0500

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Related articles
eliminating left-recursion aegis@mad.scientist.com (aegis) (2006-01-07)
Re: eliminating left-recursion rjshaw@netspace.net.au (Russell Shaw) (2006-01-08)
Re: eliminating left-recursion cdodd@acm.org (Chris Dodd) (2006-01-08)
Re: eliminating left-recursion DrDiettrich@compuserve.de (Hans-Peter Diettrich) (2006-01-09)
Re: eliminating left-recursion lojiancn@hotmail.com (jackycn) (2006-01-09)
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 From: Chris Dodd Newsgroups: comp.compilers Date: 8 Jan 2006 11:38:39 -0500 Organization: Compilers Central References: 06-01-013 Keywords: parse, LL(1) Posted-Date: 08 Jan 2006 11:38:39 EST

"aegis" <aegis@mad.scientist.com> wrote
> Given the following production:
>
> d-declarator: ID | d-declarator '[' constant ']' | '(' d-declarator ')'
> ;
>
> How can I eliminate left-recursion here? The method sketched
> out in 'Compilers, Principles, Techniques and Tools' only
> presents a very simple case...

The method in ASU is fully general -- given a production of the form:

X ::= X A | B

Trnsforiming it into

X ::= B X'
X' ::= A X' | epsilon

is equivalent, but right recursive instead of left recursive. The intuition
behind this is that the original rule is equivalent to (EBNF):

X ::= B A*

The recursive part of the original rule will expand to zero or more 'A'
components, but you eventually have to expand a B, which will go on the
beginning of the rule.

For your example:

X == d-declarator
A == '[' constant ']'
B == ID | '(' d-declarator ')'

so the replacement expansion is

d-declarator: ( ID | '(' d-declarator ')' ) d-declarator-tail
d-declarator-tail: '[' constant ']' d-declarator-tail

for which you might need to expand the first rule as

d-declarator: ID d-declarator-tail | '(' d-declarator ')' d-declarator-
tail

Chris Dodd
cdodd@acm.org

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