1. 文法 G（S）:
S -> AB
A ->Da|ε
B -> cC
C -> aADC |ε
D -> b|ε
验证文法 G（S）是不是 LL（1）文法？
SELECT( A -> Da) = FIRST(Da) = { b, a}
SELECT( A ->ε) = FOLLOW(A) = { c, b, a, # }
SELECT( C ->aADC) = FIRST(a) = { a }
SELECT( C ->ε) = FOLLOW(C) = { # }
SELECT( D ->b ) = FIRST(b) = { b }
SELECT( D ->ε) = FOLLOW(D) = { a, #}
因为
SELECT( A -> Da) ∩ SELECT( A -> ε) = { a } ≠ ∅
所以，文法G（S）不是 LL（1）文法
2.(上次作业)消除左递归之后的表达式文法是否是LL(1)文法？
E->TE’
E’->+TE’ | ε
T->FT’
T’->*FT’ | ε
F->(E) | i
SELECT( E’ -> +TE’) = FIRST(+TE’) = { + }
SELECT( E’ -> ε) = FOLLOW( E’) = { ),# }
SELECT( T’ -> *FT’) = FIRST( *FT’) = { * }
SELECT( T’ -> ε) = FOLLOW( T’) = { +,),# }
SELECT( F -> (E)) = FIRST( (E) ) = { ( }
SELECT( F -> i) = FIRST( i ) = { i }
因为
SELECT(E' -> +TE') ∩ SELECT(E' -> ɛ) = ∅
SELECT(T' -> *FT' ) ∩ SELECT(T' -> ɛ) = ∅
SELECT(F -> (E) ) ∩ SELECT(F -> i ) = ∅
所以 消除左递归后的文法是 LL(1)文法
3.接2，如果是LL(1)文法，写出它的递归下降语法分析程序代码。
E()
｛
T();
E'();
}
E'()
T()
T'()
F()
void ParseE(){
switch(lookahead){
case (,i:
ParseT();
ParseE'();
break;
default:
print("syntax error ");
exit(0);
}
}
void ParseE'(){
switch(lookahead){
case +:
MatchToken(+);
ParseT();
ParseE'();
break;
case ),#:
break;
default:
print("syntax error ");
exit(0);
}
}
void ParseT(){
switch(lookahead){
case (,i:
ParseF();
ParseT'();
break;
default:
print("syntax error ");
exit(0);
}
}
void ParseT'(){
switch(lookahead){
case *:
MatchToken(*);
ParseF();
ParseT'();
break;
case +,),#:
break;
default:
print("syntax error ");
exit(0);
}
}
void ParseF(){
switch(lookahead){
case (:
MatchToken(();
ParseE();
MatchToken());
break;
case i:
MatchToken(i);
break;
default:
print("syntax error ");
exit(0);
}
}