LR(0) generation Question

Hi I have a question about generating the LR(0) closure for an SLR(1)
Parser Table. Following is a typical example:


0) E->E,'+',T. Epsfree=t First={(,id} Follow={+,)}
1) E->T. Epsfree=t First={(,id} Follow={+,)}
2) T->T,'*',F. Epsfree=t First={(,id} Follow={+,),*}
3) T->F. Epsfree=t First={(,id} Follow={+,),*}
4) F->'(',E,')'. Epsfree=t First={(} Follow={+,),*}
5) F->'id'. Epsfree=t First={id} Follow={+,),*}
6) E_->E. Epsfree=t First={(,id} Follow={}


0:
(6,1) E_->.E;
(0,1) E->.E,'+',T;
(1,1) E->.T;
(2,1) T->.T,'*',F;
(3,1) T->.F;
(4,1) F->.'(',E,')';
(5,1) F->.'id';




1:
(6,2) E_->E.;
(0,2) E->E,.'+',T;




2:
(1,2) E->T.;
(2,2) T->T,.'*',F;




3:
(3,2) T->F.;




4:
(4,2) F->'(',.E,')';
(0,1) E->.E,'+',T;
(1,1) E->.T;
(2,1) T->.T,'*',F;
(3,1) T->.F;
(4,1) F->.'(',E,')';
(5,1) F->.'id';




5:
(5,2) F->'id'.;




6:
(0,3) E->E,'+',.T;
(2,1) T->.T,'*',F;
(3,1) T->.F;
(4,1) F->.'(',E,')';
(5,1) F->.'id';




7:
(2,3) T->T,'*',.F;
(4,1) F->.'(',E,')';
(5,1) F->.'id';




8:
(4,3) F->'(',E,.')';
? E->E,.'+',T;




9:
(0,4) E->E,'+',T.;
? T->T,.'*',F;


10:
(2,4) T->T,'*',F.;




11:
(4,4) F->'(',E,')'.;






goto[E,0]=1
goto[T,0]=2
goto[F,0]=3
goto[(,0]=4
goto[id,0]=5
goto[+,1]=6
goto[*,2]=7
goto[E,4]=8
goto[T,4]=2
goto[F,4]=3
goto[(,4]=4
goto[id,4]=5
goto[T,6]=9
goto[F,6]=3
goto[(,6]=4
goto[id,6]=5
goto[F,7]=10
goto[(,7]=4
goto[id,7]=5
goto[),8]=11




So my question is:
Wy is in Closure(8) this rule E->E,.'+',T; added. This rule appears in
Closure(1) also in Closure(9) is this rule T->T,.'*',F; added. So
perhaps i didn't understand the goto calculation. I thought that a goto
rule is added if the new goto doesn't exist in the goto list and the
rule in he closure list also doen't exist, or is this wrong?


Many Thanks
André Betz
mail@andrebetz.de

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