Re: When to use a token yacc_EOF instead of relying on 0 return value of yylex()?

Hi Peng,

> Le 11 févr. 2019 à 18:12, Peng Yu <[email protected]> a écrit :
> 
> Hi,
> 
> yacc_EOF instead of 0 is used in `yylex()`.
> 
> http://git.savannah.gnu.org/cgit/bash.git/tree/parse.y#n3257

Which is

  if (character == EOF)
    {
      EOF_Reached = 1;
      return (yacc_EOF);
    }


> The grammar explicitly relies on yacc_EOF.
> 
> http://git.savannah.gnu.org/cgit/bash.git/tree/parse.y#n376

Which is:

%left '&' ';' '\n' yacc_EOF
%left AND_AND OR_OR
%right '|' BAR_AND


> But this seems to be different from the normal usage in flex/bison.
> For example, flex has <<EOF>> (I know bash doesn't use its hand coded
> yylex()). Bison expects yylex() to just return 0 upon seeing the EOF
> of the input.

Yes, it's very different indeed: here yacc_EOF is not declared
as being EOF.

> """ (From bison manual regarding yylex)
> The null character must not be used this way, because its code is zero
> and that signifies end-of-input.
> """
> 
> So is there a reason bash parsing must the non-standard yacc_EOF? Or
> bash parsing can be solved by using a yylex() that returns 0 upon
> seeing EOF? Thanks.

I have no idea.  You'd have to study the grammar to see if there
are doing fancy things around yacc_EOF.  You could also map
declare yacc_EOF to be end-of-file to Bison, and see if things
work properly.  Insert this:

%token yacc_EOF 0

It associates yacc_EOF with the token number 0, which denotes
EOF.  However, I confess I never tried to use the symbol EOF
in grammar rules.  It's a very special token, you cannot do
anything with it.
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