秒差距错误 - 尝试似乎不起作用

发布于 2024-12-03 10:14:56 字数 1044 浏览 1 评论 0原文

我目前正在使用文本.Parsec.Expr 模块用于解析脚本语言的子集。

基本上，这种语言有两种命令：$var = expr 形式的赋值和$var = $array[$index] 形式的命令 -当然还有其他命令，但这足以解释我的问题。

我创建了一个类型 Command 来表示这一点，以及相应的解析器，其中分配的 expr 由 Parsec 的 buildExpressionParser 处理。

现在，问题来了。首先是解析代码：

main = case parse p "" "$c = $a[$b]" of
          Left err -> putStrLn . show $ err
          Right r  -> putStrLn . show $ r
    where p = (try assignment <|> command) <* eof -- (1)

整个代码（50行）粘贴在这里：链接（如果你已经解析了，应该编译安装）

问题是，解析失败，因为赋值没有成功解析，即使之前有一个try。反转解析顺序（尝试命令 <|> 赋值）可以解决问题，但在我的情况下是不可能的。

当然，我尝试进一步定位问题，在我看来，问题出在表达式解析器（由 buildExpressionParser 构建），因为如果我说 expr = failed ""< 则解析会成功。 /代码>。然而，我在秒差距来源中找不到任何可以解释这种行为的内容。

原文

I'm currently using the Text.Parsec.Expr module to parse a subset of a scripting language.

Basically, there are two kinds of commands in this language: Assignment of the form $var = expr and a Command of the form $var = $array[$index] - there are of course other commands, but this suffices to explain my problem.

I've created a type Command, to represent this, along with corresponding parsers, where expr for the assignment is handled by Parsec's buildExpressionParser.

Now, the problem. First the parsing code:

main = case parse p "" "$c = $a[$b]" of
          Left err -> putStrLn . show $ err
          Right r  -> putStrLn . show $ r
    where p = (try assignment <|> command) <* eof -- (1)

The whole code (50 lines) is pasted here: Link (should compile if you've parsec installed)

The problem is, that parsing fails, since assignment doesn't successfully parse, even though there is a try before. Reversing the parsing order (try command <|> assignment) solves the problem, but is not possible in my case.

Of course I tried to locate the problem further and it appears to me, that the problem is the expression parser (build by buildExpressionParser), since parsing succeeds if I say expr = fail "". However I can't find anything in the Parsec sources that would explain this behaviour.

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魔法少女 2024-12-10 10:14:56

您的解析器失败，因为事实上 assigment does 在这里成功消耗 $c = $a （尝试使用普通的 where p = assignment< /代码>）。那么应该有 eof （或来自 assigment 的 expr 的其余部分），因此会出现错误。当“赋值”的参数只是一个 var 时（例如 $c = $a），“命令”的开头似乎与“赋值”相同代码>）。不知道为什么你不能颠倒命令和赋值，但使这个特定示例工作的另一种方法是：

main = case parse p "" "$c = $a[$b]" of
      Left err -> putStrLn . show $ err
      Right r  -> putStrLn . show $ r
   where p = try (assignment <* eof) <|> (command <* eof)

You parser fails because in fact assigment does succeeds here consuming $c = $a (try it with plain where p = assignment). Then there is supposed to be eof (or the rest of expr from assigment) hence the error. It seems that the beggining of your 'command' is identical to your 'assignment' in the case when 'assignment''s argument is just a var (like $c = $a).
Not sure why you can't reverse command and assignment but another way to make this particular example work would be:

main = case parse p "" "$c = $a[$b]" of
      Left err -> putStrLn . show $ err
      Right r  -> putStrLn . show $ r
   where p = try (assignment <* eof) <|> (command <* eof)

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