我有一个高阶类型,并努力用它构建一些 DSL。而且我正在寻找一种方法来定义可以接受类型而无需明确指定此类型的函数。
自我描述的例子:
class Wrap[T] (val data : T)
class DSL {
def doSomething[T](x : Wrap[T]) =
println(x.data)
def <<=[T,W <: Wrap[T]](arg : W) : W = {
doSomething(arg)
arg
}
def <<-[T](arg : Wrap[T]) : Wrap[T] = {
doSomething(arg)
arg
}
def <<+[W <: Wrap[_]](arg : W) = {
doSomething(arg)
arg
}
def <<~(arg : Wrap[_]) = {
doSomething(arg)
arg
}
}
class ExtendedInt(x : Int) extends Wrap[Int](x) {
def expose() = println(data)
}
object Test {
val dsl = new DSL
val exi = new ExtendedInt(3)
val x1 = dsl <<= exi
val x2 = dsl <<- exi
val x3 = dsl <<+ exi
val x4 = dsl <<~ exi
x1.expose()
x2.expose()
x3.expose()
x4.expose()
}
我尝试了 4 种不同的方法,得到了 4 种不同的错误:
Casting.scala:15: error: no type parameters for method doSomething: (x: Wrap[T])Unit exist so that it can be applied to arguments (W)
--- because ---
argument expression's type is not compatible with formal parameter type;
found : W
required: Wrap[?T]
doSomething(arg)
^
Casting.scala:32: error: inferred type arguments [Nothing,ExtendedInt] do not conform to method <<='s type parameter bounds [T,W <: Wrap[T]]
val x1 = dsl <<= exi
^
Casting.scala:38: error: value expose is not a member of Wrap[Int]
x2.expose()
^
Casting.scala:40: error: value expose is not a member of Wrap[_$2]
x4.expose()
^
four errors found
所有错误都非常具有描述性。我不反对 scala 笨拙的类型系统和局限性。但我离我的目标还很远,我很乐意寻找另一种黑客来实现所需的功能。
还有其他我忽略的方法吗?