Question

我写了下面的Parser，目的是fail - 在空白处：

import scala.util.parsing.combinator._

object Foo extends JavaTokenParsers { 
  val wsTest = not(whiteSpace) // uses whitespace inherited from `RegexParsers`
}

为什么解析一堆空白成功？

scala> Foo.parseAll(Foo.wsTest, "          ")
res5: Foo.ParseResult[Unit] = [1.11] parsed: ()

scala> res5.successful
res6: Boolean = true

从project查看Parsers#not，我希望我的上述测试可以Failure。

  /** Wrap a parser so that its failures and errors become success and
   *  vice versa -- it never consumes any input.
   */
  def not[T](p: => Parser[T]): Parser[Unit] = Parser { in =>
    p(in) match {
      case Success(_, _)  => Failure("Expected failure", in)
      case _              => Success((), in)
    }
  }

Answer 1

JavaTokenParsers扩展了RegexParsers，RegexParsers有：

 protected val whiteSpace = """\s+""".r

 def skipWhitespace = whiteSpace.toString.length > 0

 implicit def regex(r: Regex): Parser[String] = new Parser[String] {
    ... 
    val start = handleWhiteSpace(source, offset)
    ...
 }

 protected def handleWhiteSpace(source: java.lang.CharSequence, offset: Int): Int =
   if (skipWhitespace)
     (whiteSpace findPrefixMatchOf (source.subSequence(offset, source.length))) match {
       case Some(matched) => offset + matched.end
       case None => offset
     }
   else
     offset

所以它会跳过空格（你可以通过覆盖def skipWhitespace = false来覆盖它）

因此解析器" "等于＆＃34;＆＃34;

空白试图匹配＆＃34;＆＃34;但它失败了（＆＃34;＆＃34;＆＃34; \ s +＆＃34;＆＃34;＆＃34;需要至少一个空格）而不将其转换为成功

在Parser Combinators中理解`not`

1 个答案: