【文章內(nèi)容簡介】 b], [S→?b, a/b] } = I9goto(I6, A) = {[S→A?S, a/b] , [S→?AS, a/b], [S→?b, a/b], [A→?SA, a/b], [A→?a, a/b] } = I10goto(I6, a) = I3, goto(I6, b) = I7goto(I8, S) = I5, goto(I8, A) = I6, goto(I8, a) = I3, goto(I8, b) = I7goto(I9, S) = I5, goto(I9, A) = I6, goto(I9, a) = I3, goto(I9, b) = I7goto(I10, S) = I9, goto(I10, A) = I10, goto(I10, a) = I3, goto(I10, b) = I7規(guī)范LR分析表為：actiongotoab$SA0s3s4121s3s7acc562s3s4823r4r44r2r2r25s3s7566s3/r3s7/r39107r2r28s3/r1s7/r1r1569s3/r1s7/r15610s3s7910f) 利用LR(1)項目集合并的方法構(gòu)造LALR分析表解：同心集合并：I2 10 = {[S→A?S, $/a/b] , [S→?AS, $/a/b], [S→?b, $/a/b], [A→?SA, a/b], [A→?a, a/b] }I47 = {[S→b?, $/a/b]}I89 = {[S→AS?, $/a/b], [A→S?A, a/b], [A→?SA, a/b], [A→?a, a/b], [S→?AS, a/b], [S→?b, a/b] }LALR分析表為：actiongotoab$SA0s3s4712 101s3s47acc562 10s3s47892 103r4r447r2r2r25s3s47566s3/r3s47/r3892 1089s3/r1s47/r1r156g) 利用高效構(gòu)造方法構(gòu)造LALR分析表解：LR(0)項目集的核：I0 = { S’→?S }I1 = { S’→S?, A→S?A }I2 = { S→A?S }I3 = { A→a? }I4 = { S→b? }I5 = { A→S?A }I6 = { A→SA?, S→A?S }I7 = { S→AS?, A→S?A }closure([S’→?S, ]) = { [S’→?S, ], [S→?AS, /a/b], [S→?b, /a/b], [A→?SA, a/b], [A→?a, a/b] }I0:S’→?S傳播到I1: S’→S?，I2: S→A?S，I4: S→b?自生搜索符a/b：I1: A→S?A，I2: S→A?S，I3: A→a?，I4: S→b?closure([A→S?A, ])={[A→S?A, ], [A→?SA, /a/b], [A→?a, /a/b], [S→?AS, a/b], [S→?b, a/b]}I1:A→S?A傳播到I6: A→SA?，I5: A→S?A，I3: A→a?自生搜索符a/b：I5: A→S?A，I6: S→A?S，I3: A→a?，I4: S→b?closure([S→A?S, ])={[S→A?S, ] , [S→?AS, /a/b], [S→?b, /a/b], [A→?SA, a/b], [A→?a, a/b] }I2: S→A?S傳播到I7: S→AS?，I4: S→b?自生搜索符a/b：I2: S→A?S，I7: A→S?A，I3: A→a?，I4: S→b?I5:A→S?A傳播到I6: A→SA?，I3: A→a?自生搜索符a/b：I5: A→S?A，I6: S→A?S，I3: A→a?，I4: S→b?I6: S→A?S傳播到I7: S→AS?，I2: S→A?S，I4: S→b?自生搜索符a/b：I2: S→A?S，I7: A→S?A，I3: A→a?，I4: S→b?closure([A→S?A, ])={[A→S?A, ], [A→?SA, /a/b], [A→?a, /a/b], [S→?AS, a/b], [S→?b, a/b]}I7:A→S?A傳播到I6: A→SA?，I5: A→S?A，I3: A→a?自生搜索符a/b：I5: A→S?A，I6: S→A?S，I3: A→a?，I4: S→b?計算搜索符：集合 項目搜索符初始第一遍第二遍I0S’→?S$$$I1S’→S?$$I1A→S?Aa/ba/ba/bI2S→A?Sa/b$/a/b$/a/bI3A→a?a/ba/ba/bI4S→b?a/b$/a/b$/a/bI5A→S?Aa/ba/ba/bI6A→SA?a/ba/bI6S→A?Sa/ba/ba/bI7S→AS?a/b$/a/bI7A→S?Aa/ba/ba/bLALR分析表為：actiongotoab$SA0s3s4121s3s4acc562s3s4723r4r44r2r2r25s3s4566s3/r3s4/r3727s3/r1s4/r1r156（Aho） 考慮下面文法E → E + T | TT → T F | FF → F *| a | ba） 構(gòu)造SLR分析表解：拓廣文法E39。 → EI0={ E39。 → ?E, E → ?E + T, E → ? T, T → ?T F, T → ?F, F → ?F *, F → ?a, F → ?b }goto(I0, E) = { E39。 → E?, E → E ?+ T } = I1goto(I0, T) = { E → T?, T → T ?F, F → ?F *, F → ?a, F → ?b } = I2goto(I0, F) = { T → F?, F → F ?* } = I3goto(I0, a) = { F → a? } = I4goto(I0, b) = { F → b? } = I5goto(I1, +) = { E → E + ?T, T → ?T F, T → ?F, F → ?F *, F → ?a, F → ?b } = I6goto(I2, F) = { T → T F?, F → F ?* } = I7goto(I2, a) = I4goto(I2, b) = I5goto(I3, *) = { F → F *? } = I8goto(I6, T) = { E → E + T?, T → T ?F, F → ?F *, F → ?a, F → ?b } = I9goto(I6, F) = I3goto(I6, a) = I4goto(I6, b) = I5goto(I7, *) = I8goto(I9, F) = I7goto(I9, a) = I4goto(I9, b) = I5follow(E) = { +, $ }follow(T) = { a, b, +, $ }follow(F) = { a, b, *, +, $ }SLR分析表：actiongotoab*+$ETF0s4s51231s6acc2s4s5r2r273r4r4s8r4r44r6r6r6r6r65r7r7r7r7r76s4s5937r3r3s8r3r38r5r5r5r5r59s4s5r1r17b） 構(gòu)造LALR分析表解：LR(0)項目集規(guī)范族的核I0 = { E39。 → ?E }I1 = { E39。 → E?, E → E ?+ T }I2 = { E → T?, T → T ?F }I3 = { T → F?, F → F ?* }I4 = { F → a? }I5 = { F → b? }I6 = { E → E + ?T }I7 = { T → T F?, F → F ?* }I8 = { F → F *? }I9 = { E → E + T?, T → T ?F }closure({[E39。 → ?E , ]}) = { [E39。 → ?E , ], [E → ?E + T, ], [E → ? T, +], [T → ?T F, +/a/b], [T → ?F, +/a/b], [F → ?F *, +/a/b/*], [F → ?a, +/a/b/*], [F → ?b, +/a/b/*] }傳播：I0：E39。 → ?E232。I1：E39。 → E?, I1：E → E ?+ T自生：I2：E → T?，I2：T → T ?F，I3：T → F?，I3：F → F ?*，I4：F → a?，I5：F → b?傳播：I1：E → E ?+ T232。I6：E → E + ?Tclosure({[T → T ?F, ] }) = {[T → T ?F, ], [F → ?F *, /*], [F → ?a, /*], [F → ?b, /*] }傳播：I2：T → T ?F232。I7：T → T F?，I7：F → F ?*，I4：F → a?，I5：F → b?自生：I7：F → F ?*，I4：F → a?，I5：F → b?傳播：I3：F → F ?*232。I8：F → F *?closure({[E → E + ?T, ]}) = {[E → E + ?T, ], [T → ?T F, /a/b], [T → ?F, /a/b], [F → ?F *, /a/b/*], [F → ?a, /a/b/*], [F → ?b, /a/b/*] }傳播：I6：E → E + ?T232。I9：E → E + T?，I9：T → T ?F，I3：T → F?，I3：F → F ?*，I4：F → a?，I5：F → b?自生：I9：T → T ?F，I3：T → F?，I3：F → F ?*，I4：F → a?，I5：F → b?傳播：I7：F → F ?*232。 I8：F → F *?closure({[T → T ?F, ]}) = {[T → T ?F, ], [F → ?F *, /*], [F → ?a, /*], [F → ?b, /*] }傳播：I9：T → T ?F232