Re: TODO update

[Paul Hilfinger]

 > I fail to see why $default exists here: the two valid lookaheads are $
 > and + (as denoted by [$, '+'].  So it seems to me that this state has
 > no default, and ought to trigger an error *now*, now after reduction.

Consider the following grammar, which does NOT use %glr-parser:

----------------------------------------------------------------------
%%

prog : foo | bar ;

foo : '(' glorp ')' 'X' ;

bar : '(' gleep ')' ;

glorp : 'X' ;

gleep : 'X' ;

%%
----------------------------------------------------------------------

The grammar has a R/R conflict (obviously).  The -v output is as
follows (excerpt):

----------------------------------------------------------------------
state 5

    5 glorp: 'X' .
    6 gleep: 'X' .

    ')'       reduce using rule 5 (glorp)
    ')'       [reduce using rule 6 (gleep)]
    $default  reduce using rule 5 (glorp)
----------------------------------------------------------------------

As you can see, Bison NORMALLY reports a $default reduction, even if 
(as in this case), the only valid next token is ')', already covered
in the preceding two lines.  

The $default reduction is actually used, by the way, but not in any
valid program.  If your input were

   (XX

for example, you'd get the following trace

----------------------------------------------------------------------
Starting parse
Entering state 0
Reading a token: Next token is token '(' ()
Shifting token 40 ('('), Entering state 1
Reading a token: Next token is token 'X' ()
Shifting token 88 ('X'), Entering state 5
Reducing via rule 5 (line 9), 'X'  -> glorp   <<< NOTE <<<
state stack now 0 1
Entering state 6
Reading a token: Next token is token 'X' ()
Err: parse error
Error: popping nterm glorp ()
Error: state stack now 0 1
Error: popping token '(' ()
Error: state stack now 0
----------------------------------------------------------------------

The marked line shows where the default reduction is used (the next
token is 'X', which is invalid).

Paul