100分求布尔表达式LR分析表

[fireinthehole]

我在网上找了一下，《编绎原理实践教程》这本书上有，可是我附近的书店都没这本书，请有的大哥帮下忙。
或者高手帮忙推导一下，我已经焦头烂额了……

[Kevin_qing]

%token COMP_OP IDENT LOGIC_OP1 LOGIC_OP2
%start LOGIC_EXPR
%%

factor : IDENT
| '(' COMP_EXPR ')'
;

COMP_EXPR
: factor
| COMP_EXPR COMP_OP factor
;
LOGIC_EXPR
: COMP_EXPR
| LOGIC_OP1 COMP_EXPR
| LOGIC_EXPR LOGIC_OP2 COMP_EXPR
;

COMP_OP指代 >, <, ==, >= ,<= ,!=
LOGIC_OP1指代 !
LOGIC_OP2指代 && ||


分析表


Grammar
rule 1 factor -> IDENT
rule 2 factor -> '(' COMP_EXPR ')'
rule 3 COMP_EXPR -> factor
rule 4 COMP_EXPR -> COMP_EXPR COMP_OP factor
rule 5 LOGIC_EXPR -> COMP_EXPR
rule 6 LOGIC_EXPR -> LOGIC_OP1 COMP_EXPR
rule 7 LOGIC_EXPR -> LOGIC_EXPR LOGIC_OP2 COMP_EXPR

Terminals, with rules where they appear

$ (-1)
'(' (40) 2
')' (41) 2
error (256)
COMP_OP (258) 4
IDENT (259) 1
LOGIC_OP1 (260) 6
LOGIC_OP2 (261) 7

Nonterminals, with rules where they appear

factor (9)
on left: 1 2, on right: 3 4
COMP_EXPR (10)
on left: 3 4, on right: 2 4 5 6 7
LOGIC_EXPR (11)
on left: 5 6 7, on right: 7


state 0

IDENT shift, and go to state 1
LOGIC_OP1 shift, and go to state 2
'(' shift, and go to state 3

factor go to state 4
COMP_EXPR go to state 5
LOGIC_EXPR go to state 6



state 1

factor -> IDENT . (rule 1)

$default reduce using rule 1 (factor)



state 2

LOGIC_EXPR -> LOGIC_OP1 . COMP_EXPR (rule 6)

IDENT shift, and go to state 1
'(' shift, and go to state 3

factor go to state 4
COMP_EXPR go to state 7



state 3

factor -> '(' . COMP_EXPR ')' (rule 2)

IDENT shift, and go to state 1
'(' shift, and go to state 3

factor go to state 4
COMP_EXPR go to state 8



state 4

COMP_EXPR -> factor . (rule 3)

$default reduce using rule 3 (COMP_EXPR)



state 5

COMP_EXPR -> COMP_EXPR . COMP_OP factor (rule 4)
LOGIC_EXPR -> COMP_EXPR . (rule 5)

COMP_OP shift, and go to state 9

$default reduce using rule 5 (LOGIC_EXPR)



state 6

LOGIC_EXPR -> LOGIC_EXPR . LOGIC_OP2 COMP_EXPR (rule 7)

$ go to state 14
LOGIC_OP2 shift, and go to state 10



state 7

COMP_EXPR -> COMP_EXPR . COMP_OP factor (rule 4)
LOGIC_EXPR -> LOGIC_OP1 COMP_EXPR . (rule 6)

COMP_OP shift, and go to state 9

$default reduce using rule 6 (LOGIC_EXPR)



state 8

factor -> '(' COMP_EXPR . ')' (rule 2)
COMP_EXPR -> COMP_EXPR . COMP_OP factor (rule 4)

COMP_OP shift, and go to state 9
')' shift, and go to state 11



state 9

COMP_EXPR -> COMP_EXPR COMP_OP . factor (rule 4)

IDENT shift, and go to state 1
'(' shift, and go to state 3

factor go to state 12



state 10

LOGIC_EXPR -> LOGIC_EXPR LOGIC_OP2 . COMP_EXPR (rule 7)

IDENT shift, and go to state 1
'(' shift, and go to state 3

factor go to state 4
COMP_EXPR go to state 13



state 11

factor -> '(' COMP_EXPR ')' . (rule 2)

$default reduce using rule 2 (factor)



state 12

COMP_EXPR -> COMP_EXPR COMP_OP factor . (rule 4)

$default reduce using rule 4 (COMP_EXPR)



state 13

COMP_EXPR -> COMP_EXPR . COMP_OP factor (rule 4)
LOGIC_EXPR -> LOGIC_EXPR LOGIC_OP2 COMP_EXPR . (rule 7)

COMP_OP shift, and go to state 9

$default reduce using rule 7 (LOGIC_EXPR)



state 14

$ go to state 15



state 15

$default accept

[fireinthehole]

这位兄台，能否告知一二在编程的事后如何实现？

[Kevin_qing]

根据上面的状态转换写程序撒
至于你想用分析表或者直接用switch实现就随你了
shift是移进
reduce是规约
goto是转换状态

使用的是LALR分析

[fireinthehole]

谢谢这位大哥，如有不懂的问题还要向你请教哦

[fireinthehole]

直接用switch实现的话是不是按顺序从state 0 到state 15进行？

[fireinthehole]

我碰到了一个问题，从我目前来学的，我只会用文法来推导出项目集，然后构造LR分析表，也就是那种矩阵式表格，我把上面的代码形式的分析表画成了DFA，下一步应该如何构造成分析表，构造规范LR分析表还是LALR分析表？

[Kevin_qing]

上面的是LALR分析表，已经构造好了
比如
state 0

IDENTshift, and go to state 1
LOGIC_OP1shift, and go to state 2
'(' shift, and go to state 3

factorgo to state 4
COMP_EXPRgo to state 5
LOGIC_EXPRgo to state 6

就是状态0时，输入为 IDENT 就移进 然后转换到状态1
LOGIC_OP1时移进，转换到2
'(' 转换到3

那么可以写分析表为

#define ERR ~0
#define S(x) (0x10000000|x)
#define R(x) (0x20000000|x)
#define G(x) (0x30000000|x)

unsigned long table[][6]={

//IDENT------'('------')'--------'LOGIC_OP1---------LOGIC_OP2-----CMP_OP-----FACTOR-----CMP_EXPR----LOGIC_EXPR
/*S0*/{ S(1), S(3) ERR S(2), ERR , ERR , G(4), G(5),G(6)},
//S1 .......

//S12.....
};

[Kevin_qing]

unsigned long table[][6]={
应该是
unsigned long table[][9]={

[fireinthehole]

对了，似乎and和or优先级是一样的，但是离散数学书上优先级从高到低是not and or，这个问题怎么解决？

[Kevin_qing]

你把bnf写来，我给你表

[flying_dancing]

把你的表达示(文法)写出来............

[fireinthehole]

我的课程设计题目的是这样的：
题目：布尔表达式的翻译程序（最好用C++编写）
方法：LR分析法
输出中间代码：四元式
要求：写出布尔表达式的符合分析方法要求的文法，编号分析程序后，设计若干用例，上机测试并通过所设计的分析程序。

书上给出的最初文法为：
E→E or E|E and E|not E|(E)|id relop id|id

由于要输出四元式，书上又给出文法：
E→E1 or M E2|E1 and M E2|not E1|(E1)|id1 relop id2|id
M→ε
其中M是为了便于在适当的时候执行一个语义动作，记住下一个将要产生的四元式标号
E1 E2 是算术表达式，relop为关系元算符（≤≥≠=<>）

我觉得为了定义符号优先级，于是改写文法如下，不知道是否正确：
E→E or M F|F
F→F and M G|G
G→not G|(E)|id relop id|id
M→ε



这次课程设计的题目难死了，关键在于推导出LR分析表，可是我已经尽力了，还是推不出来，请各位能够抽出一点空闲，对小弟指点一二。