某储备粮的“学习笔记” - compiler 2011-04-19T06:15:49+08:00 Typecho http://blog.gregwym.info/feed/atom/tag/compiler/ <![CDATA[Compiler各个步骤的含义]]> http://blog.gregwym.info/compiler-ge-ge-bu-zhou-de-han-yi.html 2011-04-19T06:15:49+08:00 2011-04-19T06:15:49+08:00 咳嗽di小鱼 http://blog.gregwym.info Step1: String -> Scanning (DFA) -> Tokens Scanner又叫Lexical analyzer或Lexer.

其目的在于, 将需要compile的源代码逐字读入compiler, 并将每一个符合"词汇命名规则(Lexical Syntax)"的字段转换成token存储.
换句话说, 就是一遍读, 一遍看每一个单词的拼写对不对. 对的就转成token, 拼错了就输出error.

作业中对应: A6 P4
(某些语言的compiler在Scanning之后还包含Preprocessing, 因为不在241讨论范围内, 不做解释)

Step2: Tokens -> Parsing(LL/LR) -> Intermediate Format (WLI)

Parsing又叫Syntactic Analysis.
其目的在于, 将token与token联系在一起, 并将他们的转换成符合一定规范的"中间格式", 一般是某种树状结构, 例如241中定义的WLI.

在Parsing过程中, 如果遇到不符合某种语言的"语法规则(Grammar)"时, 则输出error. 如果语法正确, 则输出对应格式.
简单说, 就是看的说的话是不是人话, 有没有缺个标点少个括号.
如果不是人话那就说明你该重新学语法去了.

作业中对应: A8 P4

Step3: Intermediate Format -> Semantic Analysis(Context-Sensitive Analysis) ->

Semantic Analysis的目的在于, 检查程序是否存在语义上的冲突. 或者说, 上下文是不是相符.

比如在C中, 如果没有declare过变量int a;, 则a = 3;就不合法.
再比如, 如果a declare为int, 则a = 'b';就不合法, 因为a不能为char.

作业中对应: A9 A10
(Optimization为代码优化, 241没有涉及, 知道即可)

Step4: -> Code Generation -> Code Fragment

将Intermediate Format转换为另一种格式, 比如MIPS或者二进制文件, 可与上一步同步进行.

作业中对应: A9 A10, A3, A4

Step5: Code Fragements -> Linking -> Executable File(Single File)

将多个compile好的多个文件链接在一起, 生成一个可执行的二进制文件(或仅生成合并以后的单个文件, 但文件本身可能无法执行)

作业中对应: A5 P1, P2

附: 09FALL 第二题答案

  1. Scanning
  2. Parsing
  3. Semantic Analysis
  4. Parsing
  5. Semantic Analysis
  6. Linking
  7. Semantic Analysis
  8. Parsing
  9. 1A@#SR%$#
  10. + foo
  11. x
]]> <![CDATA[LR(1) Parser Example]]> http://blog.gregwym.info/lr-1--parser-example.html 2011-04-19T05:24:08+08:00 2011-04-19T05:24:08+08:00 咳嗽di小鱼 http://blog.gregwym.info Terminal Symbols: 6个
BOF, EOF, id, -, (, )

Nonterminal Symbols: 3个
S, expr, term

Start Symbol:
S

Production Rules: 5个
0 S BOF expr EOF
1 expr term
2 expr expr - term
3 term id
4 term ( expr )

Total State: 11个 (0 to 11)
Total Transitions: 28个

State Symbol Action
0 BOF shift to state 6
1 ( shift to state 3
1 id shift to state 2
1 term shift to state 8
3 ( shift to state 3
3 expr shift to state 7
3 id shift to state 2
3 term shift to state 4
6 ( shift to state 3
6 expr shift to state 10
6 id shift to state 2
6 term shift to state 4
7 ) shift to state 9
7 - shift to state 1
10 - shift to state 1
10 EOF shift to state 5
2 ) reduce by rule 3
2 - reduce by rule 3
2 EOF reduce by rule 3
4 ) reduce by rule 1
4 - reduce by rule 1
4 EOF reduce by rule 1
8 ) reduce by rule 2
8 - reduce by rule 2
8 EOF reduce by rule 2
9 ) reduce by rule 4
9 - reduce by rule 4
9 EOF reduce by rule 4

String to Parse:
BOF id - ( id ) - id EOF


Parse Step:
蓝色的行是Shift, 灰色的行是Reduce
Reduce的次数取决于Production Rule RHS的长度
当前Symbol所对应的State,
State Stack
(At EOL)		 Description
S0 0 Start
BOF S6 0 6 S0 BOF S6
id S2 0 6 2 S6 id S2
R3 - 0 6 S2 - Reduce by rule #3
(replace id with term)
term S4 0 6 4 S6 term S4
R1 - 0 6 S4 - Reduce by rule #1
(replace term with expr)
expr S10 0 6 10 S6 expr S10
- S1 0 6 10 1 S10 - S1
( S3 0 6 10 1 3 S1 ( S3
id S2 0 6 10 1 3 2 S3 id S2
R3 ) 0 6 10 1 3 S2 ) Reduce by rule #3
(replace id with term)
term S4 0 6 10 1 3 4 S3 term S4
R1 ) 0 6 10 1 3 S4 ) Reduce by rule #1
(replace term with expr)
expr S7 0 6 10 1 3 7 S3 expr S7
) S9 0 6 10 1 3 7 9 S7 ) S9
R4 R4 R4 - 0 6 10 1 S9 - Reduce by rule #4
(repace "( expr )" with term)
term S8 0 6 10 1 8 S1 term S8
R2 R2 R2 - 0 6 S8 - Reduce by rule #2
(replace "expr - term" with expr)
expr S10 0 6 10 S6 expr S10
- S1 0 6 10 1 S10 - S1
id S2 0 6 10 1 2 S1 id S2
R3 EOF 0 6 10 1 S2 EOF Reduce by rule #3
(replace id with term)
term S8 0 6 10 1 8 S1 term S8
R2 R2 R2 EOF 0 6 S8 EOF Reduce by rule #2
(replace "expr - term" with expr)
expr S10 0 6 10 S6 expr S10
EOF S5 Final S5 S10 EOF S5


原题来源: http://www.student.cs.uwaterloo.ca/~cs241/parsing/lr1.html

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