基本Grako示例给出了IndexError

Question

我想开始使用Grako (3.6.6)，作为对解析器的第一次体验，我希望通过自定义语法生成一个HTML表。以下基本测试

import grako

grammar = """table = { row }+ ;
row = (cell1:cell "|" cell2:cell) "\n";
cell = /[a-z]+/ ;
"""

model = grako.genmodel("model", grammar)

ast = model.parse(
"""a | b
c | d
""", "table")
print(ast)

导致一个错误

  File "test.py", line 13, in <module>
    """, "table")
  File "grako\grammars.py", line 790, in grako.grammars.Grammar.parse (grako\grammars.c:27773)
  File "grako\grammars.py", line 97, in grako.grammars.GrakoContext.parse (grako\grammars.c:4391)
  File "grako\contexts.py", line 180, in grako.contexts.ParseContext.parse (grako\contexts.c:4313)
  File "grako\grammars.py", line 594, in grako.grammars.Rule.parse (grako\grammars.c:22253)
  File "grako\grammars.py", line 597, in grako.grammars.Rule._parse_rhs (grako\grammars.c:22435)
  File "grako\contexts.py", line 399, in grako.contexts.ParseContext._call (grako\contexts.c:10088)
  File "grako\contexts.py", line 433, in grako.contexts.ParseContext._invoke_rule (grako\contexts.c:11135)
  File "grako\grammars.py", line 435, in grako.grammars.PositiveClosure.parse (grako\grammars.c:17285)
  File "grako\contexts.py", line 695, in grako.contexts.ParseContext._positive_closure (grako\contexts.c:19286)
  File "grako\contexts.py", line 696, in grako.contexts.ParseContext._positive_closure (grako\contexts.c:19240)
  File "grako\grammars.py", line 435, in grako.grammars.PositiveClosure.parse.lambda10 (grako\grammars.c:17195)
  File "grako\grammars.py", line 547, in grako.grammars.RuleRef.parse (grako\grammars.c:20774)
  File "grako\grammars.py", line 594, in grako.grammars.Rule.parse (grako\grammars.c:22253)
  File "grako\grammars.py", line 597, in grako.grammars.Rule._parse_rhs (grako\grammars.c:22435)
  File "grako\contexts.py", line 399, in grako.contexts.ParseContext._call (grako\contexts.c:10088)
  File "grako\contexts.py", line 433, in grako.contexts.ParseContext._invoke_rule (grako\contexts.c:11135)
  File "grako\grammars.py", line 326, in grako.grammars.Sequence.parse (grako\grammars.c:11582)
  File "grako\grammars.py", line 268, in grako.grammars.Token.parse (grako\grammars.c:9463)
  File "grako\contexts.py", line 543, in grako.contexts.ParseContext._token (grako\contexts.c:13772)
  File "grako\buffering.py", line 301, in grako.buffering.Buffer.match (grako\buffering.c:9168)
IndexError: string index out of range

恰巧是partial_match = (token[0].isalpha() and token.isalnum() and self.is_name_char(self.current()) )

尽管我对解析器还不熟悉，也有点缺乏文档，但我还是想继续使用Grako。

你能帮我建立一个基本的例子来输出表格的HTML吗？

Answer 1

Grako在语法中没有正确地看到"\n"，因为在令牌中不允许换行符，并且\n是在外部的三元引号(""")字符串的上下文中计算的。如果您使用/\n/，那么一切都很好。

还请注意，如果\n将是语言的一部分，那么您可能应该编写一个@@whitespace子句，这样解析器就不会跳过字符：

@@whitespace :: /[\t ]+/

这是你的语言的正确语法：

grammar = """
@@whitespace :: /[\t ]+/
table = { row }+ ;
row = (cell1:cell "|" cell2:cell) "\\n";
cell = /[a-z]+/ ;
"""

我目前正在修补Grako，以检测和报告语法中的错误。这些更改已经在比特桶存储库中了。测试结束后我会发布一个版本。