假设我们有一个通用类型(例如Seq[E]
)和一个具体子类型(例如Seq[Int]
)。我们如何提取与抽象类型的类型参数相对应的具体类型。换句话说,我们怎么知道E -> Int
。
下面是测试所需行为的最小代码示例。该extractTypeBinding
函数将执行相关转换。
import scala.reflect.runtime.{universe => ru}
class MyFuncs
object MyFuncs {
def fn1[E](s: Seq[E]): E = ???
def fn2[K, V](m: Map[K, V]): Int = ???
}
object Scratch {
def extractTypeBinding(genType: ru.Type, typeParam: ru.Type)(concreteType: ru.Type): ru.Type = ???
def getArgTypes(methodSymbol: ru.MethodSymbol): Seq[ru.Type] =
methodSymbol.paramLists.headOption.getOrElse(Nil).map(_.typeSignature)
def main(a: Array[String]): Unit = {
// Grab the argument types of our methods.
val funcsType = ru.typeOf[MyFuncs].companion
val fn1ArgTypes = getArgTypes(funcsType.member(ru.TermName("fn1")).asMethod)
val fn2ArgTypes = getArgTypes(funcsType.member(ru.TermName("fn2")).asMethod)
val genericSeq = fn1ArgTypes.head // Seq[E]
val genericMap = fn2ArgTypes.head // Map[K, V]
// Create an extractor for the `E` in `Seq[E]`.
val seqElExtractor = extractTypeBinding(genericSeq, genericSeq.typeArgs.head) _
// Extractor for the `K` in `Map[K,V]`
val mapKeyExtractor = extractTypeBinding(genericMap, genericMap.typeArgs.head) _
// Extractor for the `V` in `Map[K,V]`
val mapValueExtractor = extractTypeBinding(genericMap, genericMap.typeArgs(1)) _
println(seqElExtractor(ru.typeOf[Seq[Int]])) // should be Int
println(seqElExtractor(ru.typeOf[Seq[Map[String, Double]]])) // should be Map[String, Double]
println(mapKeyExtractor(ru.typeOf[Map[String, Double]])) // should be String
println(mapKeyExtractor(ru.typeOf[Map[Int, Boolean]])) // should be Int
println(mapValueExtractor(ru.typeOf[Map[String, Double]])) // should be Double
println(mapValueExtractor(ru.typeOf[Map[Int, Boolean]])) // should be Boolean
}
}
基于文档字符串,似乎asSeenFrom
应该是实现的关键extractTypeBinding
。我尝试了以下实现,但返回的类型参数保持不变。
def extractTypeBinding(genType: ru.Type, typeParam: ru.Type)(concreteType: ru.Type): ru.Type =
typeParam.asSeenFrom(concreteType, genType.typeSymbol.asClass)
...
println(seqElExtractor(ru.typeOf[Seq[Int]])) // E
println(seqElExtractor(ru.typeOf[Seq[Map[String, Double]]])) // E
如果asSeenFrom
是正确的方法,正确的咒语是什么?如果没有,那该怎么办呢?
最简单的解决方案来自Dmytro Mitin的评论。
我对此进行了一些误解.typeArgs
,并通过其他一些实验得以解决。
Map[A, Map[B, C]]
只有2个类型的args(A
和Map[B, C]
)这两个现在看起来都非常直观,但是我最初做了一些愚蠢的假设。以下是我测试的修改版本,可以更清楚地实现我的初衷。
class MyFuncs
object MyFuncs {
def fn1[E](s: Seq[E]): E = ???
def fn2[K, V](m: Map[K, V]): Int = ???
}
object Scratch {
def typeArgBindings(genericType: ru.Type, concreteType: ru.Type): Map[ru.Type, ru.Type] =
// @todo consider validating both have the same base type.
genericType.typeArgs.zip(concreteType.typeArgs).toMap
def getArgTypes(methodSymbol: ru.MethodSymbol): Seq[ru.Type] =
methodSymbol.paramLists.headOption.getOrElse(Nil).map(_.typeSignature)
def main(a: Array[String]): Unit = {
// Grab the argument types of our methods.
val funcsType = ru.typeOf[MyFuncs].companion
val fn1ArgTypes = getArgTypes(funcsType.member(ru.TermName("fn1")).asMethod)
val fn2ArgTypes = getArgTypes(funcsType.member(ru.TermName("fn2")).asMethod)
val genericSeq = fn1ArgTypes.head // Seq[E]
val genericMap = fn2ArgTypes.head // Map[K, V]
println(typeArgBindings(genericSeq, ru.typeOf[Seq[Int]])) // Map(E -> Int)
println(typeArgBindings(genericSeq, ru.typeOf[Seq[Map[String, Double]]])) // Map(E -> Map[String,Double])
println(typeArgBindings(genericMap, ru.typeOf[Map[String, Double]])) // Map(K -> String, V -> Double)
println(typeArgBindings(genericMap, ru.typeOf[Map[Int, Boolean]])) // Map(K -> Int, V -> Boolean)
}
}
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