scala combinator解析器没有回溯,因为我原以为……

我有这个组合器解析器,它没有像我想象的那样回溯.我一直在将它简化为一个小例子而没有完全删除上下文.感觉像“foobar” – 例子更难阅读.我来啦：

@RunWith(classOf[JUnitRunner])
class ParserBacktrackTest extends RegexParsers with Spec with ShouldMatchers {
  override def skipWhitespace = false

  lazy val optSpace = opt(whiteSpace)
  lazy val number = """d+([.]d+)?""".r
  lazy val numWithOptSpace = number <~ optSpace

  private def litre = numWithOptSpace <~ ("litre" | "l")
  def volume = litre ^^ { case _ => "volume" }

  private def namedPieces = numWithOptSpace <~ ("pcs") ^^ { case _ => "explPcs" }
  private def implicitPieces = number ^^ { case _ => "implPcs" }
  protected def unitAmount = namedPieces | implicitPieces

  def nameOfIngredient = ".*".r

  def amount = volume | unitAmount
//  def amount = unitAmount
  protected def ingredient = (amount <~ whiteSpace) ~ nameOfIngredient

  describe("IngredientParser") {
    it("should parse volume") {
      shouldParse("1 litre lime")
    }
    it("should parse explicit pieces") {
      shouldParse("1 pcs lime")
    }
    it("should parse implicit pieces") {
      shouldParse("1 lime")
    }
  }

  def shouldParse(row: String) = {
    val result = parseAll(ingredient,row)
    result match {
      case Success(value,_) => println(value)
      case x => println(x)
    }
    result.successful should be(true)
  }
}

那么会发生的是第三次测试失败：

(volume~lime)
(explPcs~lime)
[1.4] failure: string matching regex `s+' expected but `i' found

1 lime
   ^

所以似乎升解析器消耗了l然后当它找不到任何空间时失败了.但我会认为它会回溯并尝试下一个生产规则.显然,implicitPieces解析器解析了这一行,因为如果删除前面的卷解析器(删除注释),它就会成功

(implPcs~litre lime)
(explPcs~lime)
(implPcs~lime)

为什么金额不回溯？我有什么误会？