Re: Tell a rule to shift instead of reduce?

[Ron Burk]

> because it would be wrong when looking into functions.

Is there some reason you need to treat function parameter lists
as expressions, rather than comma-separated expression lists?
Could you just follow the usual practice of:

a) give ',' and '=' the precedence you desire
b) break your expression non-terminal into two:

comma_expr
    : expr ',' expr
    | expr
expr
    : expr '+' expr'
    /* etc., but no comma operator */
    | '(' comma_expr ')'    /* comma inside parens in a func arg is an
operator */
    | func '(' arg_list ')'
arg_list
    : expr
    | expr ',' arg_list
    | %empty

(typed off the top of my head, so probably not error-free.)
This gives you a function parameter list where the parameters
are separated by commas, not comma operators.


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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