Re: TODO update

[Akim Demaille]

|  > There is something which remains unclear to me: you seem to say we
|  > should not display them as $default, but they are.  They are because
|  > conflrow and others are not used to generate .output.  
| 
| Well, no, but the change I introduced for the benefit of GLR affects
| 'actrow', and thus prevents it from being set to the 'default' marker
| (0), and therefore from being compressed away.  Take a look at the
| output you provided:

I think we are still not understanding each other.  output is
generated *before* the computation of actrow and the like.  It does
*not* use the same computation of `default' as the ``real'' automaton
does.

As I said, I agree it is a problem in the design of Bison, but i do
not want to change this now.


|Take a look at the output you provided:
|  > | State 5 contains 1 shift/reduce conflict and 2 reduce/reduce conflicts.
|  > | 
|  > | 
|  > | Grammar
|  > | 
|  > |     0 $axiom: exp $
|  > | 
|  > |     1 exp: exp '+' exp
|  > |     2    | exp '+' exp
|  > |     3    | '0'
|  > | 
|  ...
|  > | state 5
|  > | 
|  > |     1 exp: exp . '+' exp  [$, '+']
|  > |     1    | exp '+' exp .  [$, '+']
|  > |     2    | exp . '+' exp  [$, '+']
|  > |     2    | exp '+' exp .  [$, '+']
|  > | 
|  > |     '+'  shift, and go to state 4
|  > | 
|  > |     $         reduce using rule 1 (exp)
|  > |     $         [reduce using rule 2 (exp)]
|  > |     '+'       [reduce using rule 1 (exp)]
|  > |     '+'       [reduce using rule 2 (exp)]
|  > |     $default  reduce using rule 1 (exp)
| 
| This says there are conflicts on $ and '+'.  For $, it gives the two
| reductions; for '+', it gives the two possible reductions and the shift.

I'm referring to the fact that, unless I'm mistaken, there is no such
thing as a $default in GLR, so first of all, $default has nothing to
do here.

|  > The presentation of this state is very confusing.
| 
| I'm not sure why you say so.  The one potentially confusing thing is
| that the first $ rule (since there is no S/R conflict) is not
| bracketed, even though it is, in fact, treated identically to the 
| other $ rule (which is bracketed).   This is an artifact of bison's
| reporting of the OTHER sense of 'default' reduction---i.e., the
| reduction taken by default when there is a R/R conflict for the
| ordinary LALR(1) parser.

I disaagree on the reading.  What Bison wrote above is

        There are several possible actions on $: r1 and r2.
        r2 is disabled.
        There are several on +: s4, r1, r2.
        r1 and r2 are disabled.

*That's* what ``means'' Bison. `[]' means `possible according to the
grammar, but no executed at all because of conflicts', not `because of
$default'.

Play as much as you want with or without %glr, there is *no
difference*.  In the present case, it ought to be

state 5

    1 exp: exp . '+' exp  [$, '+']
    1    | exp '+' exp .  [$, '+']
    2    | exp . '+' exp  [$, '+']
    2    | exp '+' exp .  [$, '+']

    '+'  shift, and go to state 4

    $         reduce using rule 1 (exp)
    $         reduce using rule 2 (exp)
    '+'       reduce using rule 1 (exp)
    '+'       reduce using rule 2 (exp)
    $default  reduce using rule 1 (exp)

and I was, in addition, questioning the presence of this $default
guy.  Should it be:

state 5

    1 exp: exp . '+' exp  [$, '+']
    1    | exp '+' exp .  [$, '+']
    2    | exp . '+' exp  [$, '+']
    2    | exp '+' exp .  [$, '+']

    '+'  shift, and go to state 4

    $         reduce using rule 1 (exp)
    $         reduce using rule 2 (exp)
    '+'       reduce using rule 1 (exp)
    '+'       reduce using rule 2 (exp)

???