系列文章戳这里👇

1. 什么是上下文无关文法、最左推导和最右推导
2. 如何判断二义文法及消除文法二义性
3. 何时需要消除左递归
4. 什么是句柄、什么是自上而下、自下而上分析
5. 什么是LL(1)、LR(0)、LR(1)文法、LR分析表
6. LR(0)、SLR(1)、LR(1)、LALR(1)文法之间的关系
7. 编译原理第三章习题
8. 词法分析、构建DFA、上下文无关文法、LL(1)分析、提取正规式
9. 证明LL(1)、SLR(1)、LALR(1)文法
10. 翻译方案、属性栈代码

文法如下：
$$\begin{gathered} S\to (L)|a \\ L\to L,S|S \end{gathered}$$
(a) 写一个翻译方案，它输出每个 a 的嵌套深度。例如，对于句子 (a, (a, a))，输出的结果是 1 2 2。
文法符号S，L继承属性depth表示嵌套深度，则翻译方案如下：
S ′ → { S . d e p t h = 0 } S S → ( { L . d e p t h = S . d e p t h } L ) S → a { p r i n t ( S . d e p t h ) } L → { L 1 . d e p t h = L . d e p t h , S . d e p t h = L . d e p t h } L 1 , S L → { S . d e p t h = L . d e p t h } S \begin{aligned} S'&→\{S.depth=0\}S\\ S&→(\{L.depth = S.depth\}L)\\ S&→a\{print(S.depth)\}\\ L&→\{L_1.depth=L.depth,S.depth=L.depth\}L_1,S\\ L&→\{S.depth=L.depth\}S \end{aligned} S′SSLL​→{S.depth=0}S→({L.depth=S.depth}L)→a{print(S.depth)}→{L1​.depth=L.depth,S.depth=L.depth}L1​,S→{S.depth=L.depth}S​
属性栈代码，由于 属性栈上仅能存放综合属性（后文有详细介绍），所以需要引入标记非终结符P、Q、R，及其综合属性s，继承属性i，模拟继承属性的计算，则栈代码如下：
S ′ → { S . d e p t h = P . s } P S P → ϵ { P . s = 0 }                         S t a c k [ n t o p ] = 0 S → ( { Q . i = S . d e p t h , L . d e p t h = Q . s } Q L ) Q → ϵ { Q . s = Q . i + 1 }              S t a c k [ n t o p ] = S t a c k [ t o p − 1 ] + 1 S → a { p r i n t ( S . d e p t h ) }            p r i n t ( S t a c k [ t o p − 1 ] ) L → { L 1 . d e p t h = L . d e p t h , R . i = L . d e p t h , S . d e p t h = R . s } L 1 , R S R → ϵ { R . s = R . i }                     S t a c k [ n t o p ] = S t a c k [ t o p − 2 ] L → { S . d e p t h = L . d e p t h } S \begin{aligned} S'&→\{S.depth=P.s\}PS\\ P&→\epsilon\{P.s=0\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop]=0\\ S&→(\{Q.i=S.depth,L.depth = Q.s\}QL)\\ Q&→\epsilon\{Q.s=Q.i+1\}\ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop]= Stack[top-1]+1\\ S&→a\{print(S.depth)\}\ \ \ \ \ \ \ \ \ \ print(Stack[top-1])\\ L&→\{L_1.depth=L.depth,R.i=L.depth,S.depth=R.s\}L_1,RS\\ R&→\epsilon\{R.s=R.i\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop] = Stack[top-2]\\ L&→\{S.depth=L.depth\}S \end{aligned} S′PSQSLRL​→{S.depth=P.s}PS→ϵ{P.s=0}                       Stack[ntop]=0→({Q.i=S.depth,L.depth=Q.s}QL)→ϵ{Q.s=Q.i+1}            Stack[ntop]=Stack[top−1]+1→a{print(S.depth)}          print(Stack[top−1])→{L1​.depth=L.depth,R.i=L.depth,S.depth=R.s}L1​,RS→ϵ{R.s=R.i}                   Stack[ntop]=Stack[top−2]→{S.depth=L.depth}S​

(b) 写一个翻译方案，它打印出每个 a 在句子中是第几个字符。例如，当句子是 (a, (a, (a, a),(a)))时，打印的结果是 2 5 8 10 14。
使用综合属性out表示当前文法符号推出的字符总数，基础属性before表示该文法符号前有多少个字符，则翻译方案如下：
S ′ → { S . b e f o r e = 0 } S S → { L . b e f o r e = S . b e f o r e }             ( L )             { S . o u t = L . o u t + 2 } S → a { S . o u t = 1 ; p r i n t ( S . b e f o r e + 1 ) } L → { L 1 . b e f o r e = L . b e f o r e ,             S . b e f o r e = L . b e f o r e + L 1 . o u t + 1 }             L 1 , S             { L . o u t = L 1 . o u t + S . o u t + 1 } L → { S . b e f o r e = L . b e f o r e } S { L . o u t = S . o u t } \begin{aligned} S'&→\{S.before=0\}S\\ S&→\{L.before = S.before\} \\&\ \ \ \ \ \ \ \ \ \ \ (L) \\&\ \ \ \ \ \ \ \ \ \ \ \{S.out=L.out+2\}\\ S&→a\{S.out=1;print(S.before+1)\}\\ L&→\{L_1.before=L.before, \\&\ \ \ \ \ \ \ \ \ \ \ S.before=L.before+L_1.out+1\} \\&\ \ \ \ \ \ \ \ \ \ \ L_1,S \\&\ \ \ \ \ \ \ \ \ \ \ \{L.out=L_1.out+S.out+1\}\\ L&→\{S.before=L.before\}S\{L.out=S.out\} \end{aligned} S′SSLL​→{S.before=0}S→{L.before=S.before}           (L)           {S.out=L.out+2}→a{S.out=1;print(S.before+1)}→{L1​.before=L.before,           S.before=L.before+L1​.out+1}           L1​,S           {L.out=L1​.out+S.out+1}→{S.before=L.before}S{L.out=S.out}​
同理上述翻译方案栈代码如下：
S ′ → { S . b e f o r e = P . s } P S P → ϵ { P . s = 0 }                                                S t a c k [ n t o p ] = 0 S → ( { Q . i = S . b e f o r e , L . b e f o r e = Q . s }             Q L )             { S . o u t = L . o u t + 2 } Q → ϵ { Q . s = Q . i + 1 }                                     S t a c k [ n t o p ] = S t a c k [ t o p − 1 ] + 1 S → a { p r i n t ( S . b e f o r e ) }                                 p r i n t ( S t a c k [ t o p − 1 ] ) L → { L 1 . b e f o r e = L . b e f o r e ,             R . i = L . b e f o r e + L 1 . o u t + 1 ,             S . b e f o r e = R . s }             L 1 , R S             { L . o u t = L 1 . o u t + S . o u t + 1 }           S t a c k [ n t o p ] = S t a c k [ t o p − 3 ] + S t a c k [ t o p ] + 1 R → ϵ { R . s = R . i }                                            S t a c k [ n t o p ] = S t a c k [ t o p − 2 ] + S t a c k [ t o p − 1 ] + 1 L → { S . b e f o r e = L . b e f o r e } S         { L . o u t = S . o u t }                                     S t a c k [ n t o p ] = S t a c k [ t o p ] \begin{aligned} S'&→\{S.before=P.s\}PS\\ P&→\epsilon\{P.s=0\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop]=0\\ S&→(\{Q.i = S.before,L.before = Q.s\} \\&\ \ \ \ \ \ \ \ \ \ \ QL) \\&\ \ \ \ \ \ \ \ \ \ \ \{S.out=L.out+2\} \\ Q&→\epsilon\{Q.s=Q.i+1\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop]= Stack[top-1]+1\\ S&→a\{print(S.before)\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ print(Stack[top-1])\\ L&→\{L_1.before=L.before, \\&\ \ \ \ \ \ \ \ \ \ \ R.i=L.before+L_1.out+1, \\&\ \ \ \ \ \ \ \ \ \ \ S.before=R.s\} \\&\ \ \ \ \ \ \ \ \ \ \ L_1,RS \\&\ \ \ \ \ \ \ \ \ \ \ \{L.out=L_1.out+S.out+1\} \ \ \ \ \ \ \ \ \ Stack[ntop]=Stack[top-3]+Stack[top]+1 \\ R&→\epsilon\{R.s=R.i\}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop] = Stack[top-2]+Stack[top-1]+1\\ L&→\{S.before=L.before\}S \\& \ \ \ \ \ \ \ \{L.out=S.out\} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Stack[ntop]=Stack[top] \end{aligned} S′PSQSLRL​→{S.before=P.s}PS→ϵ{P.s=0}                                              Stack[ntop]=0→({Q.i=S.before,L.before=Q.s}           QL)           {S.out=L.out+2}→ϵ{Q.s=Q.i+1}                                   Stack[ntop]=Stack[top−1]+1→a{print(S.before)}                               print(Stack[top−1])→{L1​.before=L.before,           R.i=L.before+L1​.out+1,           S.before=R.s}           L1​,RS           {L.out=L1​.out+S.out+1}         Stack[ntop]=Stack[top−3]+Stack[top]+1→ϵ{R.s=R.i}                                          Stack[ntop]=Stack[top−2]+Stack[top−1]+1→{S.before=L.before}S       {L.out=S.out}                                   Stack[ntop]=Stack[top]​

属性栈代码：

• 如何在自底向上分析中计算继承属性？
  • 属性栈上仅能存放综合属性
  • 分析栈中符号的继承属性
    • 属性copy规则
      如果，$A\to XY$，有语义规则$Y.i:=X.s$，
      翻译方案可写为：
      A → X { Y . i : = X . s } Y A → X \{ Y.i := X.s \} Y A→X{Y.i:=X.s}Y
  • 从句柄下面取继承属性!

举个栗子

有如下文法与翻译方案

 D→T { L.in := T.type }  L 			
   T→int { T.type := integer }
   T→real	{ T.type := real }
   L→ { L1.in := L.in }
		  L1 , id {addtype(id.entry, L.in) }
   	L→id { addtype(id.entry, L.in) }

对于输入串：int p,q,r 分析过程如下：

引入标记非终结符模拟继承属性的计算