Re: Tell a rule to shift instead of reduce?

[Chris verBurg]

Hey Adam,

I have a couple ideas.  First, you might try updating the precedence of ','
for just that one rule:

| rvalLoop ',' rval %prec '='

(though you might need to define a precedence-level higher than both ','
and '=' for that to work right).

A second, more straightforward, idea is to shuffle around your grammar to
focus on the idea of an atomic "rval group":

body:
  recursive-expr // without rval=rval
  | rvalGroup '=' rvalGroup ;

rvalGroup:
  rval
  | rvalGroup ',' rval ;

-Chris



On Tue, Dec 17, 2013 at 12:25 AM, Adam Smalin <address@hidden>wrote:

> I still need help with this. I rewrote it maybe this will be more clear?
>
> I want to allow this
>
> var = var
> var, var = var
> var = var, var
>
> My rules are similar to the below
>
> body:
>     recursive-expr //rval = rval is here
>   | rval '=' rval ',' rvalLoop
>   | rval ',' rvalLoop '=' rval
>
> rvalLoop:
>     rval
>   | rvalLoop ',' rval
>
> The problem is '=' is higher precedence than ','. When I write "a=b,c" it
> reduce when it sees ',' making a=b an rval and rval ',' rval is invalid so
> my parser fails and "rval '=' rval ',' rvalLoop" is never used. The
> conflict file shows
>
> rval '=' rval . ',' rvalLoop
> Conflict between rule 352 and token ',' resolved as reduce (',' < '=').
>
> How do I say for this rule shift instead of reduce? I don't want to make
> ',' higher than '=' because it would be wrong when looking into functions.
> For example func(a,b=2,c) should have b=2 as an expression and not (a,b) =
> (2,c).
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