除列表之外的序列上的Scala模式匹配 [英] Scala pattern matching on sequences other than Lists

问题描述

我有以下代码递归操作列表中的每个元素

I have the following code which recursively operates on each element within a List

def doMatch(list: List[Int]): Unit = list match {
  case last :: Nil  => println("Final element.")
  case head :: tail => println("Recursing..."); doMatch(tail)
}

现在，忽略此功能可通过 filter（） 和foreach（），这个工作很好。但是，如果我尝试更改为接受任何 Seq [Int] ，则会遇到问题：

Now, ignoring that this functionality is available through filter() and foreach(), this works just fine. However, if I try to change it to accept any Seq[Int], I run into problems:

• Seq没有::，但它有+：，这是我理解基本上是一样的事情。如果我尝试匹配head +：tail，但编译器提示错误：not found：value +：'


• Nil特定于List，我不知道替换为。如果我遇到了以前的问题，我将尝试Seq（）。

这是我认为代码应该是什么样子，但它不工作：

Here is how I think the code should look, except it doesn't work:

def doMatch(seq: Seq[Int]): Unit = seq match {
  case last +: Seq() => println("Final element.")
  case head +: tail  => println("Recursing..."); doMatch(tail)
}

编辑：这么多好的答案！我接受agilesteel的答案，因为他是第一个，注意::不是我的例子中的运算符，而是一个case类，因此的区别。

So many good answers! I'm accepting agilesteel's answer as his was the first that noted that :: isn't an operator in my example, but a case class and hence the difference.

推荐答案

在Scala中有两个 :: （发音cons）。一个是在 class List 中定义的运算符，一个是 class （ List 的子类），它表示以头部和尾部为特征的非空列表。

There are two :: (pronounced cons) in Scala. One is an operator defined in class List and one is a class (subclass of List), which represents a non empty list characterized by a head and a tail.

head :: tail 是一个构造函数模式，它从 :: （head，tail）。

head :: tail is a constructor pattern, which is syntactically modified from ::(head, tail).

:: 是一个case类，有一个为它定义的提取器对象。

:: is a case class, which means there is an extractor object defined for it.

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