Question

它是用Lemon编译的，它是一个LALR（1）解析器生成器：

program ::= statement.

statement ::= ifstatement Newline.
statement ::= returnstatement Newline.

ifstatement ::= If Number A statement B.
ifstatement ::= If Number A statement B Newline Else A statement B.

returnstatement ::= Return Number.

错误消息是：

user@/tmp > lemon test.lm
test.lm:6: This rule can not be reduced.

1 parsing conflicts.

调试输出为：

State 0:
          program ::= * statement
          statement ::= * ifstatement Newline
          statement ::= * returnstatement Newline
          ifstatement ::= * If Number A statement B
          ifstatement ::= * If Number A statement B Newline Else A statement B
          returnstatement ::= * Return Number

                            If shift  10
                        Return shift  3
                       program accept
                     statement shift  13
                   ifstatement shift  12
               returnstatement shift  11

State 1:
          statement ::= * ifstatement Newline
          statement ::= * returnstatement Newline
          ifstatement ::= * If Number A statement B
          ifstatement ::= * If Number A statement B Newline Else A statement B
          ifstatement ::= If Number A statement B Newline Else A * statement B
          returnstatement ::= * Return Number

                            If shift  10
                        Return shift  3
                     statement shift  4
                   ifstatement shift  12
               returnstatement shift  11

State 2:
          statement ::= * ifstatement Newline
          statement ::= * returnstatement Newline
          ifstatement ::= * If Number A statement B
          ifstatement ::= If Number A * statement B
          ifstatement ::= * If Number A statement B Newline Else A statement B
          ifstatement ::= If Number A * statement B Newline Else A statement B
          returnstatement ::= * Return Number

                            If shift  10
                        Return shift  3
                     statement shift  8
                   ifstatement shift  12
               returnstatement shift  11

State 3:
          returnstatement ::= Return * Number

                        Number shift  14

State 4:
          ifstatement ::= If Number A statement B Newline Else A statement * B

                             B shift  15

State 5:
          ifstatement ::= If Number A statement B Newline Else * A statement B

                             A shift  1

State 6:
          ifstatement ::= If Number A statement B Newline * Else A statement B

                          Else shift  5

State 7:
      (3) ifstatement ::= If Number A statement B *
          ifstatement ::= If Number A statement B * Newline Else A statement B

                       Newline shift  6
                       Newline reduce 3   ** Parsing conflict **

State 8:
          ifstatement ::= If Number A statement * B
          ifstatement ::= If Number A statement * B Newline Else A statement B

                             B shift  7

State 9:
          ifstatement ::= If Number * A statement B
          ifstatement ::= If Number * A statement B Newline Else A statement B

                             A shift  2

State 10:
          ifstatement ::= If * Number A statement B
          ifstatement ::= If * Number A statement B Newline Else A statement B

                        Number shift  9

State 11:
          statement ::= returnstatement * Newline

                       Newline shift  16

State 12:
          statement ::= ifstatement * Newline

                       Newline shift  17

State 13:
      (0) program ::= statement *

                             $ reduce 0

State 14:
      (5) returnstatement ::= Return Number *

                     {default} reduce 5

State 15:
      (4) ifstatement ::= If Number A statement B Newline Else A statement B *

                     {default} reduce 4

State 16:
      (2) statement ::= returnstatement Newline *

                     {default} reduce 2

State 17:
      (1) statement ::= ifstatement Newline *

                     {default} reduce 1

----------------------------------------------------
Symbols:
    0: $:
    1: Newline
    2: If
    3: Number
    4: A
    5: B
    6: Else
    7: Return
    8: error:
    9: program: If Return
   10: statement: If Return
   11: ifstatement: If
   12: returnstatement: Return

Answer 1

从调试输出看一下状态7。它描述了解析器已经接受下一组令牌的情况：

    ifstatement ::= If Number A statement B *

在这种情况下，当Newline令牌出现时，解析器可以选择以下两个选项：

记住它并切换到State 6.这个转变是由你的语法中的下一个规则规定的：
```
ifstatement ::= If Number A statement B Newline Else A statement B.
```
将当前规则视为已完成并返回上级规则。这种减少是由你的语法规则规定的：
```
ifstatement ::= If Number A statement B.
```

LALR（1）解析器在这种情况下没有其他选项可以失败，因为它无法预测流中的下一个令牌。它无法预测在Newline之后的Else。

修改语法以避免这种冲突的情况。我只能添加新的行字符通常不包括在语言语法中。 Tokenizer通常将它们视为与其他空白字符类似的标记边界。

为什么这个语法会发生冲突？

1 个答案: