LL(1)文法的判断,递归下降分析程序
1.文法 G(S):
(1)S -> AB
(2)A ->Da|ε
(3)B -> cC
(4)C -> aADC |ε
(5)D -> b|ε
验证文法 G(S)是不是 LL(1)文法.
解:
First(Da)={b,a} First(ε)={ε} First(aADC)={a} First(b)={b}
Follow(A)={c,b,a,#} Follow(C)={#} Follow(D)={a,#}
SELECT(A->Da)={b,a} SELECT(A->ε)={c,b,a,#} SELECT(C->aADC)={a}
select(C->ε)={#} select(D->b)={b} select(D->c)={a,#}
SELECT(A->Da)交SELECT(A->ε)!=空
SELECT(C->aADC)交SELECT(C->ε)=空
SELECT(D->b)交SELECT(D->ε)=空
所以G(S)不是LL(1)文法。
2.法消除左递归之后的表达式文法是否是LL(1)文法?
答: Select(E' -> +TE') = First(+TE') = {+} Select(E' -> ε) = (First(ε)-{ε})∪Follow(E') = {),#} Select(T' -> *FT') = First(*FT') = {*} Select(T' -> ε) = (First(ε)-{ε})∪Follow(T') = {#,+,)}
Select(F -> (E)) = First((E)) = {(} Select(F -> i ) = First(i) = {i}
∵Select(E' -> +TE') ∩ Select(E' -> ε) = ∅ Select(T' -> *FT') ∩ Select(T' -> ε) = ∅ Select(F -> (E)) ∩ Select(F -> i ) = ∅
∴ 文法G‘(s)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()
答:
void ParseE(){ switch(lookahead){ case '(','i': ParseT(); ParseE'(); break; default: print("syntax error \n"); exit(0); } } void ParseE'(){ switch(lookahead){ case '+': MatchToken('+'); ParseT(); ParseE'(); break; case ')','#': break; default: print("syntax error \n"); exit(0); } } void ParseT(){ switch(lookahead){ case '(','i': ParseF(); ParseT'(); break; default: print("syntax error \n"); exit(0); } } void ParseT'(){ switch(lookahead){ case '*': MatchToken('*'); ParseF(); ParseT'(); break; case '+',')','#': break; default: print("syntax error \n"); exit(0); } } void ParseF(){ switch(lookahead){ case '(': MatchToken('('); ParseE(); MatchToken(')'); break; case 'i': MatchToken('i'); break; default: print("syntax error \n"); exit(0); } }