Re: LL(1) problem

Albert wrote:


<LL(1) problem with the folowing rules>


> Factor = LogicalConstant | LiteralExpression | MemberAccess | ident |
> "(" Expression ")" | "bosh".
> MemberAccess = (ClassName | "ky") "." ident [ "(" [ArgList] ")" ].
  > ClassName = [PackageName] "." ident .
  > PackageName = ident ":" ident { ":" ident }.




The problem is that, within the Factor function, you can't decide
between Factor -> ident and Factor -> MemberAccess because both
can start with "ident".


Some solutions are:
1. Use LALR parsing. It won't have the problem, but is probably a bigger
change than you want to make.


2. Note that, for MemberAccess to start with an ident, the ident must be
the start of a PackageName. You can, therefore, look up the identifer
in the symbol table to see if it is a package name (Assuming the ident
in Factor->ident is not allowed to be a package name) and make the
decision based on that.


3. Using the information from 2, the symbol following the ident will be
a ":" only if it is a PackageName. So you can simply look ahead one
symbol when lloking at an identifier and choose MemberAccess if you see
a ":", and Factor->ident otherwise. In other words, your language is
LL(2).


Cheers,
Carl.

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