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完成(2)短路计算(含回填)

创建于：2025年5月15日

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问题

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完成(2)短路计算(含回填)

思考

Completing short-circuit calculation

It seems like the user wants help completing a short-circuit calculation based on a boolean expression: "a<not b and c<d or e<f and g<h" using the given translation scheme with backpatching. The assignment involves applying two calculation methods: direct calculation and short-circuit calculation with backpatching, as the images show. I’ll focus on providing the code/steps for this given expression according to the translation scheme, making sure it properly follows the operator precedence rules.

Generating short-circuit code

The user wants to produce intermediate code for the short-circuit evaluation of a boolean expression. This will involve computing the code with proper precedence: 'not' first, then '<', followed by 'and', and 'or'. I’ll break down the expression step by step and write the corresponding translation scheme.

The grammar includes rules for the logical operators (like 'or', 'and', 'not') and relational comparisons, focusing on generating the appropriate backpatches and 3-address code with jumps. After converting the boolean expression into code, I’ll handle the syntax and output as required.

Generating short-circuit code

The evaluation involves dealing with operator precedence: 'and' binds tighter than 'or'. The tricky part is how we handle 'not b' in a relational expression like 'a < not b'. Since 'not b' is a boolean operation, I’ll compute it as a temporary value (e.g., t1 = not b), then use it in the relational comparison 'a < t1'. For short-circuit evaluation, I’ll generate code with conditional jumps, ensuring that only necessary operations are evaluated based on the structure of the boolean expression. I'll proceed by writing three-address code with labels for backpatching.

Implementing short-circuit evaluation and backpatching

To handle the boolean expression evaluation, I need to process each relational operation, like 'a < not b'. I'll compute intermediate steps like 't1 = not b' and then use it in conditional branches. Then, I’ll patch the True/False jump targets for short-circuit behavior. Each relational comparison will be mapped with specific jump commands, and line numbers will follow sequentially. After patching, labels like 'L_true' and 'L_false' will be used to replace placeholders, completing the translation with proper backpatching to enable the correct behavior in the final code.

Thought for 1m 30s

回答

下面给出 布尔表达式

a < not b and c < d or e < f and g < h

按幻灯片(2)中“短路计算(含回填)”翻译方案(带 true / false 链及 backpatch)完成的三地址代码与推导过程。为便于检阅，我把每条指令按顺序编号（从 1 开始），再说明各链及回填操作。

1 产生各基本关系式的代码

产生式语义动作生成的指令初始链
E &rarr; id_1 rop id_2emit if id_1 rop id_2 goto _ <br>emit goto _(R1) if a < t1 goto _ <br>(R2) goto _R1.true = {2}<br>R1.false = {3}
E &rarr; id_1 rop id_2(C1) if c < d goto _ <br>(C2) goto _C1.true = {4}<br>C1.false = {5}
E &rarr; id_1 rop id_2(E1) if e < f goto _ <br>(E2) goto _E1.true = {6}<br>E1.false = {7}
E &rarr; id_1 rop id_2(G1) if g < h goto _ <br>(G2) goto _G1.true = {8}<br>G1.false = {9}

其中 t1 := not b 由产生式 E &rarr; not id 先生成（编号 1），所以 R1 的条件实际比较 a 和 t1。

2 构造 “and” 节点并回填

  1. E3 = (a < not b)

  2. E4 = (c < d)

    • E1 = E3 and E4
    • 回填 E3.true &rarr; E4.codebegin → 行 2 跳向 4
    • E1.true = E4.true = {4}
    • E1.false = E3.false &cup; E4.false = {3,5}

  3. E5 = (e < f)

  4. E6 = (g < h)

    • E2 = E5 and E6
    • 回填 E5.true &rarr; E6.codebegin → 行 6 跳向 8
    • E2.true = {8}
    • E2.false = {7,9}

3 构造 “or” 节点并回填

  • 顶层 E = E1 or E2
  • 回填 E1.false &rarr; E2.codebegin → 行 3, 5 均跳向 6
  • E.true = {4, 8}
  • E.false = {7, 9}

4 最终回填 true/false 出口

设表达式求值为真时转到行 10，假时转到行 11，则

  • 回填 E.true → 10（行 4, 8）
  • 回填 E.false → 11（行 7, 9）

5 完整已回填的三地址代码

text
1:  t1 := not b
2:  if a < t1 goto 4
3:  goto 6
4:  if c < d goto 10
5:  goto 6
6:  if e < f goto 8
7:  goto 11
8:  if g < h goto 10
9:  goto 11
10:  ...   ;  真值分支 (E.true)
11:  ...   ;  假值分支 (E.false)
  • 行 2 的目标 4、行 3/5 的目标 6、行 4/8 的目标 10、行 7/9 的目标 11，均来源于各步 backpatch。
  • 该代码严格按运算符优先级实现短路求值：
    • 只有当 a < not b 为真才检测 c < d；
    • 只有当前半部分为假时才继续检测第二个 and 子式；
    • 任何一个 and 子式为真即可直接跳到真分支 10。

这样即完成了**(2) 短路计算（含回填）**的翻译。

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