During my test to get << a difference precedence then < I suspect it is
impossible because after expr < expr [<] it has no idea if it will be a <<
expr or an < expression thus I can't tell it to have a different precedence
between < and < <.
Because I can't I wrote this test grammar. When I execute `a<b< <cd<e` I
get << < < as I wanted. The issue is 2 shift reduce conflicts. I can't
understand it. It does exactly what I want (<< before <). Its has
%glr-parser at the top. Why isn't it ok shifting and reducing when it can't
complete the longer rule (mainElement3). I'm just confused all over I don't
see the conflict. How would I rewrite this to not have the conflict?
%nonassoc DOLLAR
//%left '<'
%left '*' PREC
%left '.'
%%
program: mEOS main
main: | mainLoop mEOS
mainLoop:
mainElement
| mainLoop mainElement
mainElement:
mainElement3
| mainElement '<' mainElement3 { printf("<\n"); }
mainElement3:
mainElement4
| mainElement3 '<' '<' mainElement4 { printf("<<\n"); }
mainElement4:
'$' VarName %prec DOLLAR { printf("varname\n"); }
| mainElement2
//| mainElement '*' mainElement { printf("*\n"); }
mainElement2:
VAR { printf("var %s\n", yytext); }
| Token { printf("token\n"); }
Token: '.'
VarName: VAR
| VarName '.' VAR
EOS: '\n' | ';'
mEOS: | mEOS EOS
%%
state 12
6 mainElement: mainElement3 . [$end, VAR, '.', '<', '$', '\n', ';']
9 mainElement3: mainElement3 . '<' '<' mainElement4
'<' shift, and go to state 22
'<' [reduce using rule 6 (mainElement)]
$default reduce using rule 6 (mainElement)
state 24
7 mainElement: mainElement '<' mainElement3 . [$end, VAR, '.', '<',
'$', '\n', ';']
9 mainElement3: mainElement3 . '<' '<' mainElement4
'<' shift, and go to state 22
'<' [reduce using rule 7 (mainElement)]
$default reduce using rule 7 (mainElement)
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