使用Scala宏将符号拼接在一起 [英] Splicing together symbols with Scala macros

问题描述

我正在尝试从通用Scala代码中调用FastUtil或Trove之类的专用集合库.我想实现类似

I am trying to call a specialized collections library like FastUtil or Trove from generic Scala code. I would like to implement something like

def openHashMap[@specialized K, @specialized V]: ${K}2${V}OpenHashMap = 
   new ${K}2${V}OpenHashMap()

${X}显然不是有效的Scala，而只是我用于文字替换的元符号， 这样，openHashMap[Long, Double]将返回Long2DoubleOpenHashMap，该类型将在编译时知道. Scala宏可以做到这一点吗?如果是这样，哪种口味?我知道有def宏，隐式宏，fundep实现，宏注释，类型宏(现已停产)……我认为在普通的Scala-2.10、2.10宏天堂和Scala-2.11中它们是不同的.其中哪一个适合这个?

Where the ${X} is clearly not valid Scala, but just my meta notation for text substitution, so that openHashMap[Long, Double] would return a Long2DoubleOpenHashMap the type would be known at compile time. Is this possible with Scala macros. If so, which flavour? I know there are def macros, implicit macros, fundep materialization, macro annotations, type macros (now discontinued) ... and I think these are different in plain Scala-2.10, 2.10 macro paradise and Scala-2.11. Which, if any, of these are appropriate for this?

还是还有其他可以执行此操作的功能，例如字节码操作，语言虚拟化等等?但是，我不相信我刚才提到的替代方法可以.

Or is there something else that can do this like byte code manipulation, language virtualization, ...? However, I do not believe the alternative I just mentioned can.

推荐答案

隐瞒使用准引用的想法

Stealing the idea to use quasiquotes from this answer, we first add the macro-paradise plugin:

// build.sbt
scalaVersion := "2.10.2"

resolvers += Resolver.sonatypeRepo("snapshots")

addCompilerPlugin("org.scala-lang.plugins" % "macro-paradise" % "2.0.0-SNAPSHOT" 
  cross CrossVersion.full)

然后宏如下所示:

// src/main/scala/Foo.scala
import reflect.macros.Context
import language.experimental.macros

trait Foo[A]
class IntFoo() extends Foo[Int]
class AnyFoo() extends Foo[Any]

object Foo {
  def apply[A]: Foo[A] = macro applyImpl[A]

  def applyImpl[A](c: Context)(t: c.WeakTypeTag[A]): c.Expr[Foo[A]] = {
    import c.universe._
    val aTpe    = t.tpe
    val prefix  = if (aTpe =:= typeOf[Int]) "Int" else "Any"
    val clazz   = newTypeName(s"${prefix}Foo")
    c.Expr(q"new $clazz()")
  }
}

测试用例:

// src/test/scala/Test.scala
object Test extends App {
  val fooInt = Foo[Int]
  val fooAny = Foo[Any]

  println(fooInt)
  println(fooAny)
}

---

没有宏天堂插件，您将需要像New(clazz, ???)那样手动构建树，我无法使它正常工作，所以放弃了，但是当然也有可能.

---

Without the macro-paradise-plugin, you would need to construct the tree by hand, like New(clazz, ???), I couldn't get that to work so gave up, but it certainly would be possible, too.

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