scala的pattern match的两个问题——正则表达式和List

在scala exercise上做练习，在pattern match那一章的时候遇到两个题不理解，请大家帮我看下吧。

1. 处理正则表达式

scalaval EatingRegularExpression = """Eating Alert: bear=([^,]+),\s+source=(.+)""".r //.r turns a String to a regular expression
val SittingRegularExpression = """Sitting Alert: bear=([^,]+),\s+source=(.+)""".r
val SleepingRegularExpression = """Sleeping Alert: bear=([^,]+),\s+source=(.+)""".r

def goldilocks(expr: String) = expr match {
  case (EatingRegularExpression(bear, source)) => "%s said someone's been eating my %s".format(bear, source)
  case (SittingRegularExpression(bear, source)) => "%s said someone's been sitting on my %s".format(bear, source)
  case (SleepingRegularExpression(bear, source)) => "%s said someone's been sleeping in my %s".format(bear, source)
  case _ => "what?"
}

goldilocks("Eating Alert: bear=Papa, source=porridge") should be("Papa said someone's been eating my porridge")
goldilocks("Sitting Alert: bear=Mama, source=chair") should be("Mama said someone's been sitting on my chair")

这个正则表达式的处理过程可不可以描述如下：
首先根据pattern match的部分匹配规则，知道应该选哪个case，然后将bear和source两个参数通过正则表达式过滤出来，然后根据format将参数的值放在字符串中的相应位置。

1. 处理List

val r = List(1, 2, 3) match {
  case x :: y :: Nil => y
  case _ => 0
}

r should be(0)

这个对Nil的理解不够，没想出怎么解释

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1. 正则表达式用在 pattern match 时会去匹配这个整个串，同时把group的东西提取出来
2. 首先，双冒号表示List的拼接。然后，x::y::Nil 中冒号是右结合，所以实际运行的过程是x::(y,Nil)，然后再到(x,y,Nil），这个实际上就是(x,y)。而两个元素的List跟三个元素的List匹配不上，所以到下一个匹配，即r=0