匹配类成员“错误：未找到：值 &&”

Question

我正在尝试从 Twitter Scala School 开始学习 Scala，但遇到语法错误。当我通过我的 sbt 控制台运行“基础知识继续”教程http://twitter.github.io/scala_school/basics2.html#match 中的模式匹配代码时，编译器将我返回为“错误：未找到：值 &&”。 Scala 中是否发生了一些变化，以采用在编写教程时可能起作用但现在不起作用的东西？所涉及的类是

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> 

当我对类成员进行匹配时，如何在我的案例中获取 AND 的用例？

提前致谢。我是新手。

Answer 1

如果您按值匹配，就像您的情况一样，您不仅可以使用警卫，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"
}

我发现这个更容易解析。

Answer 2

当您想使用模式测试条件时，您需要使用guard：

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

_ 表示您不关心calc 具有的具体值，而是守卫中提到的其他一些条件。

顺便说一句，可以写conjunction extractor：

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

但它不适用于您的具体情况。