如何从变量列表中创建Shapeless HMap

问题描述 投票:1回答:1

我正在使用scala-exercises.中的以下代码创建无形状的HMap,>

import shapeless.HMap

class BiMapIS[K, V]
implicit val intToString = new BiMapIS[Int, String]
implicit val stringToInt = new BiMapIS[String, Int]

val hm = HMap[BiMapIS](23 -> "foo", "bar" -> 13)

我想从可变参数创建HMap,如下所示(我的参数列表很长,所以只检查我是否可以简化代码littlebit)-

import shapeless.{HMap, HNil}
import java.util.{List => JList}
val entities: JList[(_, _)] = ???
class BiMapIS[K, V]
implicit val intToString = new BiMapIS[Int, String]
implicit val stringToInt = new BiMapIS[String, Int]
import collection.JavaConverters._
val entitiesSeq = entities.asScala.toList
val hm = HMap[BiMapIS](entitiesSeq:_*)

有什么方法可以从变量args创建HMap吗?

我在scala 2.12 https://mvnrepository.com/artifact/com.chuusai/shapeless_2.12/2.3.3中使用无毛2.33

我正在使用scala-exercises中的以下代码创建无形状的HMap。导入shapeless.HMap类BiMapIS [K,V]隐式val intToString = new BiMapIS [Int,String]隐式val stringToInt = ...

scala shapeless
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