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) syntax.
解： 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,#}
·.·SELECT(A -> Da) ∩ SELECT(A -> ε) ≠ Ø
SELECT(C -> aADC) ∩ SELECT(C -> ε) = Ø
SELECT(D -> b) ∩ SELECT(D -> ε) = Ø
. · .G (S) is not LL (1) grammar
2. After the elimination of the expression grammar is left recursive if LL (1) grammar?
Solution: E -> TE '
E '-> + TE' | e
T -> FT'
T' -> *FT' | ε
F -> (E) | i
SELECT( E' -> +TE' ) -> { + }
SELECT( E' -> ε ) -> { # }
SELECT( T' -> *FT' ) -> { * }
SELECT( T' -> ε ) -> { # }
SELECT( F -> (E) ) -> { ( }
SELECT( F -> i ) -> { i }
·. · SELECT (E '-> + TE') ∩ SELECT (E '-> ε) = O
SELECT( T' -> *FT' ) ∩ SELECT( T' -> ε ) = Ø
SELECT( F -> (E) ) ∩ SELECT( F -> i ) = Ø
· Is 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()
Analysis of program code:
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