Question

我最近遇到了scala-pickling，我正试图看看如何在项目中使用它，所以我一直在研究一个带有不可变哈希映射的case类的简单示例。在这个例子中，scala-pickling不会生成unpickler，我无法弄清楚原因。以下是REPL会议以证明问题：

scala> case class Foo(a: HashMap[Symbol,Symbol], b: HashMap[Symbol,Double], c: Symbol, d: Double)
defined class Foo

scala> val bar = Foo(new HashMap[Symbol,Symbol](), new HashMap[Symbol,Double](), 'A, 1.4)
bar: Foo = Foo(Map(),Map(),'A,1.4)

scala> val pickled = bar.pickle
pickled: scala.pickling.json.JSONPickle = 
JSONPickle({
  "tpe": "Foo",
  "c": {
    "name": "A"
  },
  "d": 1.4,
  "a": {

  },
  "b": {

  }
})

scala> val unpickled = pickled.unpickle[Foo]
<console>:18: error: Cannot generate an unpickler for Foo. Recompile with -Xlog-implicits for details
   val unpickled = pickled.unpickle[Foo]

任何人都可以指出我做错了什么吗？或者scala-pickling有什么问题吗？

编辑：实际上，当我生成一个只有一个符号属性的类（我将发布另一个REPL会话）时，似乎也会发生同样的情况。在scala-pickling中有没有一种特殊的方法来处理符号？

scala> case class Foo(symb: Symbol)
defined class Foo

scala> val foo = Foo('A)
foo: Foo = Foo('A)

scala> val pick = foo.pickle
pick: scala.pickling.json.JSONPickle = 
JSONPickle({
  "tpe": "Foo",
  "symb": {
    "name": "A"
  }
})

scala> val unpick = pick.unpickle[Foo]
<console>:17: error: Cannot generate an unpickler for Foo. Recompile with -Xlog-implicits for details
   val unpick = pick.unpickle[Foo]

Answer 1

我知道这是一篇旧帖子，但请尝试

import scala.pickling._, Defaults._, json._

无法为具有不可变HashMaps的案例类生成unpickler

1 个答案: