给定模块类，获取模块符号，scala宏 [英] Get the module symbol, given I have the module class, scala macro

问题描述

我正在尝试使用宏构建简单的类型类 IsEnum [T] .

I'm trying to build a simple typeclass IsEnum[T] using a macro.

如果 T ，我使用 knownDirectSubclasses 来获取所有直接子类，确保 T 是密封特征，并且所有子类都是大小写对象(使用 subSymbol.asClass.isModuleClass&& subSymbol.asClass.isCaseClass ).

I use knownDirectSubclasses to get all the direct subclasses if T, ensure T is a sealed trait, and that all subclasses are of case objects (using subSymbol.asClass.isModuleClass && subSymbol.asClass.isCaseClass).

现在，我正在尝试使用子类引用的case对象构建 Seq .

Now I'm trying to build a Seq with the case objects referred by the subclasses.

它正在工作，并且有一种解决方法:

It's working, using a workaround:

  Ident(subSymbol.asInstanceOf[scala.reflect.internal.Symbols#Symbol].sourceModule.asInstanceOf[Symbol])

但是我从其他一些问题中复制了它，似乎很老套和错误.为什么行得通?有没有更清洁的方法来实现这一目标?

But I copied that from some other question, yet it seems hacky and wrong. Why does that work? and is there a cleaner way to achieve that?

推荐答案

在2.13中，您可以实现 scala.ValueOf

In 2.13 you can materialize scala.ValueOf

val instanceTree = c.inferImplicitValue(appliedType(typeOf[ValueOf[_]].typeConstructor, subSymbol.asClass.toType))
q"$instanceTree.value"

树会有所不同

sealed trait A
object A {
  case object B extends A
  case object C extends A
}

//scalac: Seq(new scala.ValueOf(A.this.B).value, new scala.ValueOf(A.this.C).value)

但在运行时仍为 Seq(B，C).

在2.12版本中 shapeless.Witness 可以代替 ValueOf

In 2.12 shapeless.Witness can be used instead of ValueOf

val instanceTree = c.inferImplicitValue(appliedType(typeOf[Witness.Aux[_]].typeConstructor, subSymbol.asClass.toType))
q"$instanceTree.value"

//scalac: Seq(Witness.mkWitness[App.A.B.type](A.this.B.asInstanceOf[App.A.B.type]).value, Witness.mkWitness[App.A.C.type](A.this.C.asInstanceOf[App.A.C.type]).value)

libraryDependencies += "com.chuusai" %% "shapeless" % "2.4.0-M1" // in 2.3.3 it doesn't work

---

在Shapeless中，他们使用了

---

In Shapeless they use kind of

subSymbol.asClass.toType match {
  case ref @ TypeRef(_, sym, _) if sym.isModuleClass => mkAttributedQualifier(ref)
}

https://github.com/milessabin/shapeless/blob/master/core/src/main/scala/shapeless/singletons.scala#L230

或者在我们的情况下简单

or in our case simply

mkAttributedQualifier(subSymbol.asClass.toType)

，但是他们的 mkAttributedQualifier 也使用向下转换来编译器内部，并且获得的树类似于 Seq(A.this.B，A.this.C).

but their mkAttributedQualifier also uses downcasting to compiler internals and the tree obtained is like Seq(A.this.B, A.this.C).

也

Ident(subSymbol.companionSymbol)

似乎可以工作(树是 Seq(B，C))，但不推荐使用 .companionSymbol (在scaladocs中，其写为​​可能为模块类返回意外结果").即用于对象).

seems to work (tree is Seq(B, C)) but .companionSymbol is deprecated (in scaladocs it's written "may return unexpected results for module classes" i.e. for objects).

与 @MateuszKubuszok 在其库枚举中使用的方法类似.a>您也可以尝试

Following approach similar to the one used by @MateuszKubuszok in his library enumz you can try also

val objectName = symbol.fullName
c.typecheck(c.parse(s"$objectName"))

，树是 Seq(App.A.B，App.A.C).

最后，如果您对树 Seq(B，C)(而不是更复杂的树)感兴趣，似乎可以替换

Finally, if you're interested in the tree Seq(B, C) (and not some more complicated tree) it seems you can replace

Ident(subSymbol.asInstanceOf[scala.reflect.internal.Symbols#Symbol].sourceModule.asInstanceOf[Symbol])

更常规

Ident(subSymbol.owner.info.decl(subSymbol.name.toTermName))

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