Re: Change to shift instead of reduce

[Akim Demaille]

Le 1 juin 2013 à 20:49, Adam Smalin <address@hidden> a écrit :

> I have another thread but here is a simple version of my problem
> 
> My text input is
> a.b c.d
> What I would like is
> varname varname
> My grammar is
> %glr-parser
> //...
> mainLoop:
>      mainElement            %dprec 5
>    | mainLoop mainElement    %dprec 4
> mainElement:
>      '$' VarName    %dprec 6    { printf("varname\n"); }
>    | VAR             %dprec 3    { printf("var\n"); }
>    | Token         %dprec 2    { printf("token\n"); }
> 
> Token: '.'            %dprec 1
> VarName: VAR         %dprec 7
>    | VarName '.' VAR %dprec 8
> 
> The conflict file shows
>    6 mainElement: '$' VarName .  [$end, VAR, '.', '$', '\n', ';']
>   11 VarName: VarName . '.' VAR
> 
>    '.'  shift, and go to state 18
> 
>    '.'       [reduce using rule 6 (mainElement)]
>    $default  reduce using rule 6 (mainElement)
> 
> Now I see $default  reduce. I'd like it to shift. I threw in a bunch of
> precedence hoping something would trigger a shift instead of a reduce but
> no. What can I do?

You have a competition between r6, which wants to be reduced,
and token '.', which wants to be shifted.  r6 is "tagged" by
'$', its rightmost token (use %prec to use another associativity
/ precedence carrier).  So you have a competition between "$" and ".".
To make the latter win, several choices:

- use precedence (e.g., %nonassoc '$' %nonassoc '.')

- use associativity (to promote shift, make it right
  associative: %right '$' '.').

HTH.