类成员"error: not found: value &&"的匹配

Question

我正试图从Twitter Scala学校开始学习Scala，但我被语法错误绊倒了。当我通过我的sbt控制台从"基础继续"教程http://twitter.github.io/scala_school/basics2.html#match运行模式匹配代码时，编译器将我带回来"错误:未找到:值&&"。Scala是否做了一些改变，以采用在编写教程时可能有效但现在不起作用的东西?所涉及的类有

class Calculator(pBrand: String, pModel: String) {
  /**
   * A constructor
   */
  val brand: String = pBrand
  val model: String = pModel
  val color: String = if (brand.toUpperCase == "TI") {
    "blue"
  } else if (brand.toUpperCase == "HP") {
    "black"
  } else {
    "white"
  }
  // An instance method
  def add(m: Int, n: Int): Int = m + n
}
class ScientificCalculator(pBrand: String, pModel: String) extends Calculator(pBrand: String, pModel: String) {
  def log(m: Double, base: Double) = math.log(m) / math.log(base)
}
class EvenMoreScientificCalculator(pBrand: String, pModel: String) extends ScientificCalculator(pBrand: String, pModel: String) {
  def log(m: Int): Double = log(m, math.exp(1))
}

我的repl看起来像这样…

bobk-mbp:Scala_School bobk$ sbt console
[info] Set current project to default-b805b6 (in build file:/Users/bobk/work/_workspace/Scala_School/)
[info] Starting scala interpreter...
[info] 
Welcome to Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_17).
Type in expressions to have them evaluated.
Type :help for more information.
...
scala> def calcType(calc: Calculator) = calc match {
     |   case calc.brand == "hp" && calc.model == "20B" => "financial"
     |   case calc.brand == "hp" && calc.model == "48G" => "scientific"
     |   case calc.brand == "hp" && calc.model == "30B" => "business"
     |   case _ => "unknown"
     | }
<console>:9: error: not found: value &&
         case calc.brand == "hp" && calc.model == "20B" => "financial"
                                 ^
<console>:10: error: not found: value &&
         case calc.brand == "hp" && calc.model == "48G" => "scientific"
                                 ^
<console>:11: error: not found: value &&
         case calc.brand == "hp" && calc.model == "30B" => "business"
                                 ^
scala> 

当我在类成员上进行匹配时，我如何获得与在我的案例上的用例?

提前感谢。我是新手

Answer 1

如果按值匹配，就像您的情况一样，您不仅可以使用保护符，还可以坚持使用普通模式匹配:

def calcType(calc: Calculator) = (calc.brand, calc.model)  match {
     case ("hp", "20B") => "financial"
     case ("hp", "48G") => "scientific"
     case ("hp", "30B") => "business"
     case _             => "unknown"
}

Answer 2

当你想用一个模式测试一个条件时，你需要使用一个保护:

calc match {
  case _ if calc.brand == "hp" && calc.model == "20B" => "financial"
  ...
}

对于_，您表示您不关心calc的具体值，而是关心警卫中提到的其他一些条件。

顺便说一句，可以写一个连接提取器:

object && {
  def unapply[A](a: A) = Some((a, a))
}

但是它在你的具体情况下不起作用