LL(1)文法例题

已知文法G(S)如下回答以下问题S → AA → B AA → A | εB → C BB → B | εC → (A) |i(1) 计算每个非终结符的 FIRST 和 FOLLOW 集合(2)证明文法是 LL (1) 的(3)构造预测分析表(4)构造递归下降分析程序(5)按顺序写出对句子ii进行预测分析时分析栈的状态4.递归下降分析程序void S() {A();}void A() {B();A1();}void A1() {if (lookahead ) {match();A();} else {// A → εreturn;}}void B() {C();B1();}void B1() {if (lookahead ) {match();B();} else {// B → εreturn;}}void C() {if (lookahead () {match(();A();match());} else if (lookahead i) {match(i);} else {error();}}void match(char t) {if (lookahead t) {lookahead nextToken();} else {error();}}lookahead nextToken();S();if (lookahead #) {printf(分析成功);} else {error();}