Question

我正试图从Twitter Scala学校开始使用Scala，但我在语法错误方面遇到了麻烦。当我通过我的sbt控制台运行“基础继续”教程http://twitter.github.io/scala_school/basics2.html#match中的模式匹配代码时，编译器会回复“错误：未找到：值＆amp;＆amp;”。在Scala中有什么改变，以便在编写教程时可能有用，但现在不起作用？涉及的课程是

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))
}

我的副本看起来像这样...

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>

当我在班级成员上进行匹配时，如何在我的案例中获得AND的用例？

提前致谢。我是新来的。

Answer 1

如果您按值匹配，就像在您的情况下一样，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

如果要使用模式测试条件，则需要使用guard：

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

使用_表示您不需要具体值calc，而是保护中提到的其他条件。

是的，可以写一个conjunction extractor：

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

但它不适用于你的具体案例。

在班级成员上匹配“错误：找不到：值＆amp;＆amp;”

2 个答案: