将字符串匹配为 Seq[Char] 的模式 [英] Pattern matching a String as Seq[Char]

问题描述

在 Scala 中，可以通过将字符串视为 Seq[Char] 来根据字符串的单个字符来制定模式.

In Scala it is possible formulate patterns based on the invididual characters of a string by treating it as a Seq[Char].

A Tour of Scala 中提到了此功能的示例

这是那里使用的示例代码:

This is the example code used there:

object RegExpTest1 extends Application {
 def containsScala(x: String): Boolean = {
   val z: Seq[Char] = x
   z match {
      case Seq('s','c','a','l','a', rest @ _*) =>
                println("rest is "+rest)
                true
      case Seq(_*) =>
                false
   }
 }

}

我遇到的问题是代码片段的第三行:

The problem I have with this is the third line of the snippet:

val z: Seq[Char] = x

为什么需要这种演员表?在所有情况下(包括模式匹配)，字符串不应该像 Seq[Char] 一样吗?但是，如果没有这种转换，代码片段将无法工作.

Why is this sort of cast necessary? Shouldn't a String behave like a Seq[Char] under all circumstances (which would include pattern matching)? However, without this conversion, the code snippet will not work.

推荐答案

不能 100% 确定这是否正确，但我的直觉是，如果没有这个显式转换，您将模式匹配 java.lang.String，这不是您想要的.

Not 100% sure if this is correct, but my intuition says that without this explicit cast you would pattern match against java.lang.String, which is not what you want.

显式转换强制 Scala 编译器使用 Predef.stringWrapper 隐式转换；因此，正如 RichString 扩展了 Seq[Char]，你可以进行模式匹配，就好像字符串是一个字符序列.

The explicit cast forces the Scala compiler to use Predef.stringWrapper implicit conversion; thus, as RichString extends Seq[Char], you are able to do a pattern match as if the string were a sequence of characters.

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