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?
answer:
FIRST集：
FIRST(A) = { b , a , ε}
FIRST(C) = { a , ε}
FIRST(D) = { b , ε}
FOLLOW集：
FOLLOW (A) = { c , b , a , ε}
FOLLOW (C) = { # }
FOLLOW (D) = { a , #}
SELECT set:
SELECT( A -> Da ) = FIRST( Da ) = { b , a }
SELECT( A -> Da) = FIRST(Da) = { b, a }
SELECT( A -> ε) = FOLLOW( A) = { c, b, a, # }
SELECT( C -> aADC) = FIRST( aADC) = { a }
SELECT( C -> ε) = FOLLOW(C) = { # }
SELECT( D -> b) = FIRST(b) = { b }
SELECT( D -> ε ) =FOLLOW(D) = { a, # }
Since SELECT (A -> Da) ∩ SELECT (A -> ε) = {a} ≠ ∅, so grammar G (S) is not LL (1) syntax.
2. ( last job ) to eliminate left after the recursive expression grammar whether it is LL (1) grammar?
1. Elimination left recursive following grammar, the symbol string analyzing i * i + i.
Seeking FIRST set respectively, FOLLOW sets, and set SELECT
E -> E+T | T
T -> T*F | F
F -> (E) | i
answer:
Eliminate left recursion:
E -> TE '
E '-> + TE' | e
T -> FT'
T' -> *FT' | ε
F -> (E) | i
FIRST集：
FIRST(E) = { ( , i }
FIRST (E ') = {+, e}
FIRST(T) = { ( , i }
FIRST(T') = { * , ε }
FIRST(F) = { ( , i }
FOLLOW集：
FOLLOW(E) = { ) , # }
FOLLOW(E') = { ) , # }
FOLLOW(T) = { + , ) ,#}
FOLLOW(T') = {+ , ) ,#}
FOLLOW(F) = {* , + , ) ,#}
SELECT set:
SELECT (E -> TE') = FIRST(TE') = { ( , i }
SELECT(E' -> +TE') = FIRST(+TE') = { + }
SELECT(E' -> ε) = FIRST(ε) - {ε} U FOLLOW(E') = FOLLOW(E') = { ) , # }
SELECT(T -> FT') = FIRST(FT') = { ( , i }
SELECT(T' -> *FT') = FIRST(*FT') = { * }
SELECT(T' -> ε) = FIRST(ε) - {ε} U FOLLOW(T') = FOLLOW(T') = { + , ) ,# }
SELECT(F -> (E)) = FIRST((E)) = { ( }
SELECT(F -> i) = FIRST(i) = { i }
Since SELECT (E '-> + TE') ∩ SELECT (E '-> ε) = ∅,
SELECT(T' -> *FT') ∩ SELECT(T' -> ε) = ∅，
SELECT(F -> (E)) ∩ SELECT(F -> (E)) = ∅，
Therefore, the grammar 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()
answer:
void ParseE(){
if(lookhead =='(' || lookhead == 'i' ){
Frset ();
ParseE'();
}
else{
printf("syntx error\n");
exit(0);
}
}
void ParseE'(){
switch(lookahead){
case '+':
Match Token ( '+');
Frset ();
ParseE'();
break;
case ')' , '#':
break;
default:
printf("syntx error\n");
exit(0);
}
}
void ParseT(){
if(lookhead == '(' || lookhead == 'i' ){
ParseF();
Frset '();
}
else{
printf("syntx error\n");
exit(0);
}
}
void ParseT'(){
switch(lookahead){
case '*':
Match Token ( '*');
ParseF();
Frset '();
break;
case '+' , ')' , '#':
break;
default:
printf("syntx error\n");
exit(0);
}
}
void ParseF(){
switch(lookahead){
case '(':
Match Token ( '(');
ParseE();
Match Token ( ')');
break;
case 'i':
Match Token ( 'in');
break;
default:
printf("syntx error\n");
exit(0);
}
}