C&C++   发布时间:2022-04-03  发布网站:大佬教程  code.js-code.com
大佬教程收集整理的这篇文章主要介绍了解析 – 如何解决三元表达式(a?b:c)和“可能”表达式(a?)之间的LR(1)语法歧义?大佬教程大佬觉得挺不错的,现在分享给大家,也给大家做个参考。
我创建了一个语法,其精简版本如下:

(0) exp1: ternary;
(1) exp1: exp2;
(2) ternary: exp2 "?" exp1 ":" exp1;
(3) exp2: exp2 "+" exp3;
(4) exp2: exp3;
(5) exp3: maybe;
(6) exp3: "1";
(7) maybe: exp3 "?";

我相信这种语言是明确的,应该是LR可解析的. (如果我错了,请告诉我!)

但是,当我尝试为这个语法生成一个LR(1)解析器时,我得到了shift / reduce冲突,因为当解析器看到带有lookahead的exp3时,它不知道是要移位还是减少:

Conflicts in state 3:
    Reduction using rule 4: exp2:  exp3 · | "?"
    Shift to state 6

Conflicts in state 9:
    Reduction using rule 3: exp2:  exp2 "+" exp3 · | "?"
    Shift to state 6

Conflicts in state 13:
    Reduction using rule 4: exp2:  exp3 · | "?"
    Shift to state 16

Conflicts in state 20:
    Reduction using rule 4: exp2:  exp3 · | "?"
    Shift to state 23

Conflicts in state 25:
    Reduction using rule 3: exp2:  exp2 "+" exp3 · | "?"
    Shift to state 23

Conflicts in state 28:
    Reduction using rule 3: exp2:  exp2 "+" exp3 · | "?"
    Shift to state 16

有没有合理的方法让我使这种语言LR(1) – 可分解(没有冲突)?

或者GLR(或LR(2)?)是我这样的语言的唯一现实选择?
(或者我甚至错误地认为语言首先是明确的?)

作为参,我生成的模糊状态机如下(其中♦是EOF):

State 0:
    exp1:  · ternary | {♦} → shift 1
    ternary:  · exp2 "?" exp1 ":" exp1 | {♦} → shift 2
    exp2:  · exp2 "+" exp3 | {"?","+"} → shift 2
    exp2:  · exp3 | {"?","+"} → shift 3
    exp3:  · maybe | {"?","+"} → shift 4
    exp3:  · "1" | {"?","+"} → shift 5
    maybe:  · exp3 "?" | {"?","+"} → shift 3

State 1:
    exp1:  ternary · | {♦} → reduce 0

State 2:
    ternary:  exp2 · "?" exp1 ":" exp1 | {♦} → shift 7
    exp2:  exp2 · "+" exp3 | {"?","+"} → shift 8

State 3:
    exp2:  exp3 · | {"+"} → reduce 4
    exp2:  exp3 · | {"?"} → reduce 4 shift 6
    maybe:  exp3 · "?" | {"?","+"} → reduce 4 shift 6

State 4:
    exp3:  maybe · | {"?","+"} → reduce 5

State 5:
    exp3:  "1" · | {"?","+"} → reduce 6

State 6:
    maybe:  exp3 "?" · | {"?","+"} → reduce 7

State 7:
    exp1:  · ternary | {":"} → shift 10
    exp1:  · exp2 | {":"} → shift 11
    ternary:  · exp2 "?" exp1 ":" exp1 | {":"} → shift 11
    ternary:  exp2 "?" · exp1 ":" exp1 | {♦} → shift 12
    exp2:  · exp2 "+" exp3 | {"?",":","+"} → shift 11
    exp2:  · exp3 | {"?","+"} → shift 13
    exp3:  · maybe | {"?","+"} → shift 14
    exp3:  · "1" | {"?","+"} → shift 15
    maybe:  · exp3 "?" | {"?","+"} → shift 13

State 8:
    exp2:  exp2 "+" · exp3 | {"?","+"} → shift 9
    exp3:  · maybe | {"?","+"} → shift 9

State 9:
    exp2:  exp2 "+" exp3 · | {"+"} → reduce 3
    exp2:  exp2 "+" exp3 · | {"?"} → reduce 3 shift 6
    maybe:  exp3 · "?" | {"?","+"} → reduce 3 shift 6

State 10:
    exp1:  ternary · | {":"} → reduce 0

State 11:
    exp1:  exp2 · | {":"} → reduce 1
    ternary:  exp2 · "?" exp1 ":" exp1 | {":"} → shift 26
    exp2:  exp2 · "+" exp3 | {"?","+"} → shift 27

State 12:
    ternary:  exp2 "?" exp1 · ":" exp1 | {♦} → shift 17

State 13:
    exp2:  exp3 · | {":","+"} → reduce 4
    exp2:  exp3 · | {"?"} → reduce 4 shift 16
    maybe:  exp3 · "?" | {"?","+"} → reduce 4 shift 16

State 14:
    exp3:  maybe · | {"?","+"} → reduce 5

State 15:
    exp3:  "1" · | {"?","+"} → reduce 6

State 16:
    maybe:  exp3 "?" · | {"?","+"} → reduce 7

State 17:
    exp1:  · ternary | {♦} → shift 1
    exp1:  · exp2 | {♦} → shift 18
    ternary:  · exp2 "?" exp1 ":" exp1 | {♦} → shift 18
    ternary:  exp2 "?" exp1 ":" · exp1 | {♦} → shift 19
    exp2:  · exp2 "+" exp3 | {♦,"?","+"} → shift 18
    exp2:  · exp3 | {♦,"+"} → shift 20
    exp3:  · maybe | {♦,"+"} → shift 21
    exp3:  · "1" | {♦,"+"} → shift 22
    maybe:  · exp3 "?" | {♦,"+"} → shift 20

State 18:
    exp1:  exp2 · | {♦} → reduce 1
    ternary:  exp2 · "?" exp1 ":" exp1 | {♦} → shift 7
    exp2:  exp2 · "+" exp3 | {♦,"+"} → shift 24

State 19:
    ternary:  exp2 "?" exp1 ":" exp1 · | {♦} → reduce 2

State 20:
    exp2:  exp3 · | {♦,"+"} → reduce 4
    exp2:  exp3 · | {"?"} → reduce 4 shift 23
    maybe:  exp3 · "?" | {♦,"+"} → reduce 4 shift 23

State 21:
    exp3:  maybe · | {♦,"+"} → reduce 5

State 22:
    exp3:  "1" · | {♦,"+"} → reduce 6

State 23:
    maybe:  exp3 "?" · | {♦,"+"} → reduce 7

State 24:
    exp2:  exp2 "+" · exp3 | {♦,"+"} → shift 25
    exp3:  · maybe | {♦,"+"} → shift 25

State 25:
    exp2:  exp2 "+" exp3 · | {♦,"+"} → reduce 3
    exp2:  exp2 "+" exp3 · | {"?"} → reduce 3 shift 23
    maybe:  exp3 · "?" | {♦,"+"} → reduce 3 shift 23

State 26:
    exp1:  · ternary | {":"} → shift 10
    exp1:  · exp2 | {":"} → shift 11
    ternary:  · exp2 "?" exp1 ":" exp1 | {":"} → shift 11
    ternary:  exp2 "?" · exp1 ":" exp1 | {":"} → shift 29
    exp2:  · exp2 "+" exp3 | {"?","+"} → shift 13

State 27:
    exp2:  exp2 "+" · exp3 | {"?","+"} → shift 28
    exp3:  · maybe | {"?","+"} → shift 28

State 28:
    exp2:  exp2 "+" exp3 · | {":","+"} → reduce 3
    exp2:  exp2 "+" exp3 · | {"?"} → reduce 3 shift 16
    maybe:  exp3 · "?" | {"?","+"} → reduce 3 shift 16

State 29:
    ternary:  exp2 "?" exp1 · ":" exp1 | {":"} → shift 30

State 30:
    exp1:  · ternary | {":"} → shift 10
    exp1:  · exp2 | {":"} → shift 11
    ternary:  · exp2 "?" exp1 ":" exp1 | {":"} → shift 11
    ternary:  exp2 "?" exp1 ":" · exp1 | {":"} → shift 31
    exp2:  · exp2 "+" exp3 | {"?","+"} → shift 13

State 31:
    ternary:  exp2 "?" exp1 ":" exp1 · | {":"} → reduce 2

解决方法

我认为这可能是一个优先问题.当解析器查看类似这样的内容时,会发生冲突:

a + b ? c : d

在解析器看到b的时候?而且正在看c,它无法决定是否需要

>减少b ?,以便它将解析a(b?)形式的表达式,然后从那里继续,或者
>减少一个b,以便它将解析形式的表达式(a b)? c:d

我认为这里的挑战是,在一个案例中,?具有非常低的优先级(当用作三元运算符时),而在另一种情况下,它具有非常高的优先级(当用作一元运算符时).但是,如果您确实以这种方式分配优先级,我认为解析器可能能够消除这些情况的歧义.

希望这可以帮助!

大佬总结

以上是大佬教程为你收集整理的解析 – 如何解决三元表达式(a?b:c)和“可能”表达式(a?)之间的LR(1)语法歧义?全部内容,希望文章能够帮你解决解析 – 如何解决三元表达式(a?b:c)和“可能”表达式(a?)之间的LR(1)语法歧义?所遇到的程序开发问题。

如果觉得大佬教程网站内容还不错,欢迎将大佬教程推荐给程序员好友。

本图文内容来源于网友网络收集整理提供,作为学习参考使用,版权属于原作者。
如您有任何意见或建议可联系处理。小编QQ:384754419,请注明来意。