Hug*_*ira 6 scala pattern-matching
有人可以给我一个更简单的解决方案来解决以下代码(在给定结构的情况下展开整数列表0xFC :: len :: payload :: ... :: 0x0A :: 0x0D):
object Payload {
def unapply(z: List[Int]): Option[List[Int]] = if (z.length == z.head + 1) Some(z tail) else None
}
object EndToken {
def unapply(z: List[Int]): Option[List[Int]] = z.reverse match {
case 0x0D :: 0x0A :: tail => Some(tail.reverse)
case _ => None
}
}
object Message {
def unapply(z: List[Int]): Option[List[Int]] = z match {
case 0xFC :: EndToken(x) => Some(x)
case _ => None
}
}
object Main extends App {
val x = List(0xFC, 0x03, 0x01, 0x02, 0x03, 0x0A, 0x0D)
x match {
case Message(Payload(payload)) => println (payload)
case _ => println("No match")
}
}
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就像是:
object Message {
def unapply(z: List[Int]): Option[List[Int]] = z match {
case 0xFC :: Payload(x) :: 0x0A :: 0x0D => Some(x)
case _ => None
}
}
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但是,当然,::期待元素,而不是列表,所以它不起作用......
这是我的解决方案(尽管重新阅读后我认为这就像丹尼尔的解决方案)。它基于中缀操作模式,其中模式op(p, q)相同p op q。
运算符以 开头,:具有相同的优先级,以::结尾,与:右侧关联。(len, payload) :!: tail是相同的:!:((len, payload), tail)。基于长度的有效负载提取的实现有点复杂,但主要是因为我只想遍历列表一次。
object :!: {
type LengthPayload = (Int, List[Int]) // (len, payload)
// returns ((len, payload), unparsed)
def unapply(z: List[Int]): Option[(LengthPayload, List[Int])] = {
if (z == Nil) None
else {
val len = z.head
// use ListBuffer to traverse the list only once
val buf = collection.mutable.ListBuffer[Int]()
def take(l: Int, list: List[Int]): Option[(LengthPayload, List[Int])] = {
list match {
case Nil if l > 0 => None
case _ if l == 0 => Some((len, buf.toList), list)
case _ => buf += list.head; take(l - 1, list.tail)
}
}
take(len, z.tail)
}
}
}
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然后消息变得更简单(视觉上):
object Message {
def unapply(z: List[Int]): Option[List[Int]] = z match {
case 0xFC :: (len, payload) :!: 0x0A :: 0x0D :: Nil => Some(payload)
case _ => None
}
}
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结果:
val x = List(0xFC, 0x03, 0x01, 0x02, 0x03, 0x0A, 0x0D)
x match {
case Message(payload) => println(payload)
case _ => println("No match")
}
// List(1, 2, 3)
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