我一直盯着自己的这个问题看,我想这可能是一个非常愚蠢的问题。但我必须吞下我的骄傲。
我有这个组合器解析器,它不会像我想象的那样回溯。我一直在将其简化为一个小例子,而没有完全删除上下文。感觉像“foobar”-示例更难阅读。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)
}
}
那么第三次测试失败了:
(volume~lime)
(explPcs~lime)
[1.4] failure: string matching regex `\s+' expected but `i' found
1 lime
^
所以似乎litre-parser
消耗了 l 然后它在找不到任何空间时失败了。但我原以为它会回溯并尝试下一个生产规则。显然implicitPieces
解析器会解析这一行,因为如果我删除前面的卷解析器(删除注释),它会成功
(implPcs~litre lime)
(explPcs~lime)
(implPcs~lime)
为什么不amount
回溯?我有什么误解?