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系列文章戳这里👇
什么是上下文无关文法、最左推导和最右推导如何判断二义文法及消除文法二义性何时需要消除左递归什么是句柄、什么是自上而下、自下而上分析什么是LL(1)、LR(0)、LR(1)文法、LR分析表LR(0)、SLR(1)、LR(1)、LALR(1)文法之间的关系编译原理第三章习题词法分析、构建DFA、上下文无关文法、LL(1)分析、提取正规式证明LL(1)、SLR(1)、LALR(1)文法翻译方案、属性栈代码【运行时环境】什么是活动记录、 活动记录与汇编代码的关系
编译原理第三章习题
系列文章戳这里👇Homework 412345
Homework 4
3.1 3.1 3.1文法: S → ( L ) ∣ a L → L , S ∣ S F o l l o w ( S ) = { , , ) , $ } F o l l o w ( L ) = { , , ) } S→(L)|a\\ L→L,S|S\\ Follow(S)=\{,,),\$\}\\Follow(L)=\{,,)\} S→(L)∣aL→L,S∣SFollow(S)={,,),$}Follow(L)={,,)} 1( 1 ) (1) (1)习题 3.8 3.8 3.8,其中 ( b ) (b) (b)给出递归下降语法分析程序。 3.8 : 3.8: 3.8: ( a ) (a) (a)消除习题 3.1 3.1 3.1文法的左递归 消除左递归得到 S → ( L ) ∣ a L → S L ′ L ′ → , S L ′ ∣ ϵ S→(L)|a\\L→SL'\\L'→,SL'|\epsilon S→(L)∣aL→SL′L′→,SL′∣ϵ ( b ) (b) (b)为 ( a ) (a) (a)的文法构造预测分析器 F i r s t ( S ) = { ( , a } F i r s t ( L ) = F i r s t ( S ) = { ( , a } F i r s t ( L ′ ) = { , , ϵ } F o l l o w ( L ) = { ) } F o l l o w ( L ′ ) = F o l l o w ( L ) = { ) } F o l l o w ( S ) = F i r s t ( L ′ ) − { ϵ } + F o l l o w ( L ) + F o l l o w ( L ′ ) + { $ } = { , , ) , $ } \begin{aligned} First(S)&=\{(,a\}\\ First(L)&=First(S)=\{(,a\}\\ First(L')&=\{,,\epsilon\}\\ Follow(L)&=\{)\}\\ Follow(L')&=Follow(L)=\{)\}\\ Follow(S)&=First(L')-\{\epsilon\}+Follow(L)+Follow(L')+\{\$\}=\{,,),\$\}\\ \end{aligned} First(S)First(L)First(L′)Follow(L)Follow(L′)Follow(S)={(,a}=First(S)={(,a}={,,ϵ}={)}=Follow(L)={)}=First(L′)−{ϵ}+Follow(L)+Follow(L′)+{$}={,,),$} ( ( ( ) ) ) , , , a a a $ $ $ S S S S → ( L ) S→(L) S→(L) S → a S→a S→a L L L L → S L ′ L→SL' L→SL′ L → S L ′ L→SL' L→SL′ L ′ L' L′ L ′ → ϵ L'→\epsilon L′→ϵ L ′ → , S L ′ L'→,SL' L′→,SL′递归下降程序: #include #include #include const int N = 1e3; char str[N]; int ind = 0; void S();// S->(L)|a; void L();// L->SX void X();// X-> ,SX|ε int main() { int len; int m; printf("请输入要分析的串:"); scanf("%s", str); len = strlen(str); str[len] = '#'; str[len + 1] = '\0'; S(); if (str[ind] == '#') printf("正确语句!\n"); else { printf("分析失败!\n"); } return 0; } void L() {// L->SX S(); X(); } void X() {// X-> ,SX|ε if (str[ind] == ',') { ind++; S(); X(); } } void S() {// S->(L)|a; if (str[ind] == 'a') { ind++; } else if (str[ind] == '(') { ind++; L(); if (str[ind] == ')') { ind++; } else { printf("分析失败!\n"); exit(0); } } else { printf("分析失败!\n"); exit(0); } }2 ( 2 ) (2) (2)习题 3.11 3.11 3.11,并描述该文法产生的语言。 3.11 3.11 3.11: 构造下面文法的 L L ( 1 ) LL(1) LL(1)分析表 S → a B S ∣ b A S ∣ ϵ A → b A A ∣ a B → a B B ∣ b F i r s t ( S ) = { a , b , ϵ } F i r s t ( A ) = { a , b } F i r s t ( B ) = { a , b } F o l l o w ( S ) = { $ } F o l l o w ( A ) = { a , b , $ } F o l l o w ( B ) = { a , b , $ } S→aBS|bAS|\epsilon\\ A→bAA|a\\ B→aBB|b\\ \ \\ \ \\ \begin{aligned} First(S)&=\{a,b,\epsilon\}\\ First(A)&=\{a,b\}\\ First(B)&=\{a,b\}\\ Follow(S)&=\{\$\}\\ Follow(A)&=\{a,b,\$\}\\ Follow(B)&=\{a,b,\$\}\\ \end{aligned} S→aBS∣bAS∣ϵA→bAA∣aB→aBB∣b First(S)First(A)First(B)Follow(S)Follow(A)Follow(B)={a,b,ϵ}={a,b}={a,b}={$}={a,b,$}={a,b,$} a a a b b b $ $ $ S S S S → a B S S→aBS S→aBS S → b A S S→bAS S→bAS S → ϵ S→\epsilon S→ϵ A A A A → a A→a A→a A → b A A A→bAA A→bAA B B B B → a B B B→aBB B→aBB B → b B→b B→b该文法产生一个空串或者 a , b a,b a,b数量相等的 a b ab ab串。 3( 3 ) (3) (3)为习题 3.1 3.1 3.1文法构造 S L R ( 1 ) SLR(1) SLR(1)分析表。 拓广文法 S ′ → S S → ( L ) S → a L → L , S L → S \begin{aligned} &S'→S\\ &S→(L)\\ &S→a\\ &L→L,S\\ &L→S \end{aligned}\\ S′→SS→(L)S→aL→L,SL→S 该文法的 L R ( 0 ) LR(0) LR(0)项目集规范族: I 0 : S ′ → ⋅ S S → ⋅ ( L ) S → ⋅ a I 1 : S ′ → S ⋅ I 2 : S → ( ⋅ L ) L → ⋅ L , S L → ⋅ S S → ⋅ ( L ) S → ⋅ a I 3 : S → a ⋅ I 4 : S → ( L ⋅ ) L → L ⋅ , S I 5 : L → S ⋅ I 6 : S → ( L ) ⋅ I 7 : L → L , ⋅ S S → ⋅ ( L ) S → ⋅ a I 8 : L → L , S ⋅ \begin{aligned} I_0:&\\ &S'→·S\\ &S→·(L)\\ &S→·a\\ \\ \\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned} I_1:&\\ &S'→S·\\ \\ \\ \\ \\ \end{aligned} \ \ \ \ \ \ \ \begin{aligned} I_2:&\\ &S→(·L)\\ &L→·L,S\\ &L→·S\\ &S→·(L)\\ &S→·a\\ \end{aligned}\\ \ \\ \begin{aligned} I_3:&\\ &S→a·\\ \\ \\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned}\ \ \ \ I_4:&\\ &S→(L·)\\ &L→L·,S\\ \ \\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned} I_5:&\\ &L→S·\\ \\ \\ \end{aligned}\\ \begin{aligned}\ \ \ \ \ \ \ \ I_6:&\\ &S→(L)·\\ \\ \\ \end{aligned} \begin{aligned}\ \ \ \ \ \ \ \ \ I_7:&\\ &L→L,·S\\ &S→·(L)\\ &S→·a\\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned} I_8:&\\ &L→L,S·\\ \\ \\ \end{aligned}\\ \begin{aligned} \end{aligned}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I0:S′→⋅SS→⋅(L)S→⋅a I1:S′→S⋅ I2:S→(⋅L)L→⋅L,SL→⋅SS→⋅(L)S→⋅a I3:S→a⋅ I4: S→(L⋅)L→L⋅,S I5:L→S⋅ I6:S→(L)⋅ I7:L→L,⋅SS→⋅(L)S→⋅a I8:L→L,S⋅ S L R ( 1 ) SLR(1) SLR(1)分析表: ( ( ( ) ) ) a a a , , , $ $ $ S S S L L L 0 0 0 s 2 s2 s2 s 3 s3 s3 1 1 1 1 1 1 a c c acc acc 2 2 2 s 2 s2 s2 s 3 s3 s3 5 5 5 4 4 4 3 3 3 r 3 r3 r3 r 3 r3 r3 r 3 r3 r3 4 4 4 s 5 s5 s5 s 6 s6 s6 5 5 5 r 5 r5 r5 r 5 r5 r5 6 6 6 r 2 r2 r2 r 2 r2 r2 r 2 r2 r2 7 7 7 s 2 s2 s2 s 3 s3 s3 8 8 8 8 8 8 r 4 r4 r4 r 4 r4 r4 4( 4 ) (4) (4)习题 3.21 、 3.22 、 3.25 3.21、3.22、3.25 3.21、3.22、3.25。 3.21 3.21 3.21: ( a ) (a) (a)证明下面文法是 L L ( 1 ) LL(1) LL(1)文法,但不是 S L R ( 1 ) SLR(1) SLR(1)文法。 S → A a A b ∣ B b B a A → ϵ B → ϵ S→AaAb|BbBa\\ A→\epsilon\\ B→\epsilon S→AaAb∣BbBaA→ϵB→ϵ 根据 L L ( 1 ) LL(1) LL(1)文法定义, F i r s t ( A a A a ) = { a } , F i r s t ( B b B b ) = { b } First(AaAa)=\{a\},First(BbBb)=\{b\} First(AaAa)={a},First(BbBb)={b} F i r s t ( A a A a ) ∩ F i r s t ( B b B b ) = ϕ First(AaAa)∩First(BbBb)=\phi First(AaAa)∩First(BbBb)=ϕ,所以该文法是 L L ( 1 ) LL(1) LL(1)文法。 根据 S L R ( 1 ) SLR(1) SLR(1)分析,存在如下状态 I : S → ⋅ A a A b S → ⋅ B b B a A → ϵ ⋅ B → ϵ ⋅ \begin{aligned} I:&\\ &S→·AaAb\\ &S→·BbBa\\ &A→\epsilon·\\ &B→\epsilon· \end{aligned} I:S→⋅AaAbS→⋅BbBaA→ϵ⋅B→ϵ⋅ 此时,由于 F o l l o w ( A ) = F o l l o w ( B ) = { a , b } Follow(A)=Follow(B)=\{a,b\} Follow(A)=Follow(B)={a,b},所以若输入符号为 a o r b a\ or\ b a or b则无法判断是将 ϵ \epsilon ϵ归约成 A A A还是 B B B,出现归约-归约冲突,所以不是 S L R ( 1 ) SLR(1) SLR(1)文法。 ( b ) (b) (b) 证明所有 L L ( 1 ) LL(1) LL(1)文法都是 L R ( 1 ) LR(1) LR(1)文法。 证 考虑证明:若一个文法不是 L R ( 1 ) LR(1) LR(1)文法,则它一定不是 L L ( 1 ) LL(1) LL(1)文法。 若一个文法不是 L R ( 1 ) LR(1) LR(1)文法,则它存在分析动作的冲突(移进-规约和归约-归约冲突),也就是说,该文法存在二义性的最左推导,于是不满足 L L ( 1 ) LL(1) LL(1)文法的要求,所以该文法不是 L L ( 1 ) LL(1) LL(1)文法。即说明所有 L L ( 1 ) LL(1) LL(1)文法都是 L R ( 1 ) LR(1) LR(1)文法。 另外,LL(1)文法和LR(1)文法对比,两者都是从左往右看,并且只看下一个符号,再进行相 应的动作。满足LL(1)文法的语言,对它进行LL(1)分析的每一步实际上也对应着LR(1)分析的每一步。但是,LL(1)文法看到一个符号后,就提前选择对应的产生式进行推导,而LR(1)文法有了句柄的概念后,拥有了更多的信息,可以在看到一个符号时,根据搜索符进行移进或规约动作(它可以看完了句柄再选择产生式),也就是说,LR(1)文法是大于LL(1)文法的,即所有LL(1)文法都是LR(1)文法。 3.22 3.22 3.22: 证明下面文法是 L A L R ( 1 ) LALR(1) LALR(1)文法,但不是 S L R ( 1 ) SLR(1) SLR(1)文法。 S → A a ∣ b A c ∣ d c ∣ b d a A → d S→Aa|bAc|dc|bda\\ A→d\\ S→Aa∣bAc∣dc∣bdaA→d 该文法存在一个状态, [ S → d ⋅ c ] , [ A → d ⋅ ] [S→d·c],[A→d·] [S→d⋅c],[A→d⋅]出现在同一个项目集中,因为 F o l l o w ( A ) = { a , c } Follow(A)=\{a,c\} Follow(A)={a,c},所以当 输入符号为 c c c时,无法判断进行 [ S → d ⋅ c ] [S→d·c] [S→d⋅c]移进还是 [ A → d ⋅ ] [A→d·] [A→d⋅]规约,出现移进-归约冲突,所以不是 S L R ( 1 ) SLR(1) SLR(1)文法。 除上述冲突,同理还存在一个 [ S → b d ⋅ a ] , [ A → d ⋅ ] [S→bd·a],[A→d·] [S→bd⋅a],[A→d⋅]的移进-归约冲突,但这两个冲突,在规范 L R ( 1 ) LR(1) LR(1)情况下不存在,因为只有输入符号为 a a a时,才进行 d d d到 A A A的归约操作,所以此文法是 L R ( 1 ) LR(1) LR(1)文法,且此文法的规范 L R ( 1 ) LR(1) LR(1)项目集中任意项目的搜索符只能为 $ o r a $\ or\ a $ or a,所以不存在同心的 L R ( 1 ) LR(1) LR(1)项目集,所以此文法也是 L A L R ( 1 ) LALR(1) LALR(1)文法。 3.25 3.25 3.25: 一个非 L R ( 1 ) LR(1) LR(1)的文法如下: L → M L b ∣ a M → ϵ L→MLb|a\\ M→\epsilon\\ L→MLb∣aM→ϵ 请给出所有含移进-归约冲突的规范 L R ( 1 ) LR(1) LR(1)项目集,以说明该文法确实不是 L R ( 1 ) LR(1) LR(1)的。 以下项目集存在移进-归约冲突,即输入符号为 a a a时,无法判断进行移进 a a a还是进行 ϵ \epsilon ϵ规约,所以该文法不是 L R ( 1 ) LR(1) LR(1)的。 I 0 : ⟶ M L ′ → ⋅ L , $ L → ⋅ M L b , $ L → ⋅ a , $ M → ⋅ , a I 2 : ⟶ M L → M ⋅ L b , $ L → ⋅ M L b , b L → ⋅ a , b M → ⋅ , a I 5 : L → M ⋅ L b , b L → ⋅ M L b , b L → ⋅ a , b , b M → ⋅ , a \begin{aligned} I_0:&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \stackrel{M}{\longrightarrow}\\ &L'→·L,\ \ \$\\ &L→·MLb,\ \ \$\\ &L→·a,\ \ \$\\ &M→·,\ \ a\\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned} I_2:&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \stackrel{M}{\longrightarrow}\\ &L→M·Lb,\ \ \$\\ &L→·MLb,\ \ b\\ &L→·a,\ \ b\\ &M→·,\ \ a\\ \end{aligned} \ \ \ \ \ \ \ \begin{aligned} I_5:&\\ &L→M·Lb,\ \ b\\ &L→·MLb,\ \ b\\ &L→·a,b,\ \ b\\ &M→·,\ \ a\\ \end{aligned} I0: ⟶ML′→⋅L, $L→⋅MLb, $L→⋅a, $M→⋅, a I2: ⟶ML→M⋅Lb, $L→⋅MLb, bL→⋅a, bM→⋅, a I5:L→M⋅Lb, bL→⋅MLb, bL→⋅a,b, bM→⋅, a 5( 5 ) (5) (5)为以下文法构造 L R ( 1 ) LR(1) LR(1)分析表: S → a S S → A A → a A b A → \begin{aligned} &S→ a S\\ &S→ A\\ &A→ a A b\\ &A→ \end{aligned} S→aSS→AA→aAbA→ 拓广文法 S ′ → S S → a S S → A A → a A b A → ϵ \begin{aligned} &S'→ S\\ &S→ a S\\ &S→ A\\ &A→ a A b\\ &A→\epsilon \end{aligned} S′→SS→aSS→AA→aAbA→ϵ 该文法的 L R ( 1 ) LR(1) LR(1)项目集规范族为 I 0 : S ′ → ⋅ S , $ S → ⋅ a S , $ S → ⋅ A , $ A → ⋅ a A b , $ A → ⋅ ϵ , $ I 1 : S ′ → S ⋅ , $ I 2 : S → a ⋅ S , $ A → a ⋅ A b , $ S → ⋅ a S , $ S → ⋅ A , $ A → ⋅ a A b , b / $ A → ⋅ ϵ , b / $ I 3 : S → A ⋅ , $ I 4 : S → a S ⋅ , $ I 5 : A → a A ⋅ b , $ S → A ⋅ , $ I 6 : S → a ⋅ S , $ S → ⋅ a S , $ S → ⋅ A , $ A → a ⋅ A b , b / $ A → ⋅ a A b , b / $ A → ⋅ ϵ , b / $ I 7 : A → a A b ⋅ , $ I 8 : S → A ⋅ , $ A → a A ⋅ b , b / $ I 9 : A → a A b ⋅ , b / $ \begin{aligned} I_0:&\\ &S'→·S,\ \ \ \$\\ &S→·aS,\ \ \ \$\\ &S→·A,\ \ \ \$\\ &A→·aAb,\ \ \$\\ &A→·\epsilon,\ \ \$ \\\\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned} I_1:&\\ &S'→S·,\ \ \ \$\\ \\\\\\\\\\ \end{aligned} \ \ \ \ \ \ \ \begin{aligned} I_2:&\\ &S→a·S,\ \ \ \$\\ &A→a·Ab,\ \ \$\\ &S→·aS,\ \ \ \$\\ &S→·A,\ \ \ \$\\ &A→·aAb,\ \ b/\$ \\ &A→·\epsilon,\ \ b/\$\\ \end{aligned}\\ \ \\ \begin{aligned} I_3:&\\ &S→A·,\ \ \ \$\\ \ \\\\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned}\ \ \ \ I_4:&\\ &S→aS·,\ \ \ \$\\ \ \\\\ \end{aligned}\ \ \ \ \ \ \ \begin{aligned} \ \ \ I_5:&\\ &A→aA·b,\ \ \$\\ &S→A·,\ \ \ \$\\ \\ \end{aligned}\\ \begin{aligned}\ I_6:&\\ &S→a·S,\ \ \ \$\\ &S→·aS,\ \ \ \$\\ &S→·A,\ \ \ \$\\ &A→a·Ab,\ \ b/\$ \\ &A→·aAb,\ \ b/\$ \\ &A→·\epsilon,\ \ b/\$\\ \end{aligned} \begin{aligned}\ \ \ \ \ \ I_7:&\\ &A→aAb·,\ \ \ \$\\ \\\\\\\\\\ \end{aligned} \begin{aligned}\ \ \ \ \ \ \ \ I_8:&\\ &S→A·,\ \ \ \$\\ &A→aA·b,\ \ b/\$ \\ \\\\\\\\ \end{aligned}\\ \ \ \ \begin{aligned}\\ I_9:&\\ &A→aAb·,\ \ b/\$ \\ \end{aligned} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I0:S′→⋅S, $S→⋅aS, $S→⋅A, $A→⋅aAb, $A→⋅ϵ, $ I1:S′→S⋅, $ I2:S→a⋅S, $A→a⋅Ab, $S→⋅aS, $S→⋅A, $A→⋅aAb, b/$A→⋅ϵ, b/$ I3: S→A⋅, $ I4: S→aS⋅, $ I5:A→aA⋅b, $S→A⋅, $ I6:S→a⋅S, $S→⋅aS, $S→⋅A, $A→a⋅Ab, b/$A→⋅aAb, b/$A→⋅ϵ, b/$ I7:A→aAb⋅, $ I8:S→A⋅, $A→aA⋅b, b/$ I9:A→aAb⋅, b/$ L R ( 1 ) LR(1) LR(1)分析表为: a a a b b b $ $ $ S S S A A A 0 0 0 s 2 s2 s2 r 4 r4 r4 1 1 1 3 3 3 1 1 1 a c c acc acc 2 2 2 s 6 s6 s6 r 4 r4 r4 r 4 r4 r4 4 4 4 5 5 5 3 3 3 r 2 r2 r2 4 4 4 r 1 r1 r1 5 5 5 s 7 s7 s7 r 2 r2 r2 6 6 6 r 4 r4 r4 7 7 7 s 6 s6 s6 r 4 r4 r4 r 4 r4 r4 4 4 4 8 8 8 8 8 8 s 9 s9 s9 r 2 r2 r2 9 9 9 r 3 r3 r3 r 3 r3 r3 |
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