scala 如何将 List 转换为 ListBuffer？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/14443085/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to convert List to ListBuffer?
提问by Dan
Is there any way to efficiently do this, perhaps through toBuffer or to methods? My real problem is I'm building a List off a parser, as follows:
有没有办法有效地做到这一点，也许通过 toBuffer 或方法？我真正的问题是我正在从解析器构建一个列表，如下所示：
lazy val nodes: Parser[List[Node]] = phrase(( nodeA | nodeB | nodeC).*)
But after building it, I want it to be a buffer instead - I'm just not sure how to build a buffer straight from the parser.
但是在构建它之后，我希望它成为一个缓冲区 - 我只是不确定如何直接从解析器构建一个缓冲区。
回答by Régis Jean-Gilles
toindeed does the trick, and it is pretty trivial to use:
to确实可以解决问题，而且使用起来非常简单：
scala> val l = List(1,2,3)
l: List[Int] = List(1, 2, 3)
scala> l.to[ListBuffer]
res1: scala.collection.mutable.ListBuffer[Int] = ListBuffer(1, 2, 3)
Works in scala 2.10.x
适用于 Scala 2.10.x
For scala 2.9.x, you can do:
对于 Scala 2.9.x，您可以执行以下操作：
scala> ListBuffer.empty ++= l
res1: scala.collection.mutable.ListBuffer[Int] = ListBuffer(1, 2, 3)
