1. grammar G (S):
（1）S -> AB
（2）A ->Da|ε
（3）B -> cC
（4）C -> aADC |ε
（5）D -> b|ε
Verify grammar G (S) is not LL (1) grammar?
solution:
FIRST集：
FIRST（Da）={b,a}
FIRST (e) = {e}
FIRST（aADC）={a}
FIRST（b）={b}
FOLLOW集：
FOLLOW(A)=FIRST(B)+FIRST(B)+FIRST(B)+FOLLOW(C)={c,a,b,#}
FOLLOW(C)={#}
FOLLOW(D)={a,#}
SELECT set:
SELECT(A->Da)={b,a}
SELECT(A->ε)={c,b,a,#}
SELECT(C->aADC)={a}
SELECT(C->ε)={#}
SELECT(D->b)={b}
SELECT(D->ε)={a,#}
because:
SELECT(A->Da)∩SELECT(A->ε)≠∅
This grammar therefore not LL (1) syntax.
2. Whether the expression grammar after the elimination of left recursion method is LL (1) grammar?
A: Yes, LL (1) grammar.
3. The connection 2, if it is LL (1) grammar, write its recursive descent parser code.
E()
｛T();
E'();
}
E'()
T()
T'()
F()
answer:
void ParseE()｛
Frset ();
ParseE'();
}
void ParseT() {
ParseF();
Frset '();
}
void ParseE'() {
switch(lookahead):
case +:
Match Token (+);
Frset ();
ParseE'();
break;
case #:
break;
case ):
break;
default:
printf('synax error!\n');
exit(0);
}
void ParseF() {
switch(lookahead):
case (:
Match Token (();
ParseE()；
MatchToken())；
break;
case i:
Match Token (in);
break;
default:
printf('synax error!\n');
exit(0);
}
void ParseT'()
{
switch(lookahead):
case *:
ParseF();
Match Token (*);
Frset '();
break;
case #:
break;
case ):
break;
case +:
break;
default:
printf('synax error!\n');
exit(0);
}
Experiment 4. Add a lexical analyzer, parser can form a run, analyzing any input symbol string is not a valid expression.