我在O
输出类型的错误类型推断上苦苦挣扎了很长时间。为什么scalac看到Int
而不是(Int,String)
?:
trait Request[I,+O,C[_]]
case class Get[I, O, C[_]](storeName: String, arg: C[I]) extends Request[I,(I,O),C]
object Question {
val get: Request[Int,(Int,String), List] = Get("play", List(1))
}
[error] found : com.viagraphs.idb.Get[Int,Int,List]
[error] required: com.viagraphs.idb.Request[Int,(Int, String),List]
[error] val get: Request[Int,(Int,String), List] = Get("play", List(1))
请忽略W,R,ValidKey
类型类,我猜它们在这里无关紧要。
完全相同的情况发生在这种情况下:
case class Append[I : W, O : R : ValidKey](storeName: String, arg: List[I]) extends Request[I,(O,I),List]
object Question {
val get: Request[Int,(Int,String), List] = Get("play", List(1))
}
val append: Request[String,(Int,String), List] = Append("play", List("foo"))
[error] found : com.viagraphs.idb.Append[String,String]
[error] required: com.viagraphs.idb.Request[String,(Int, String),List]
[error] val append: Request[String,(Int,String), List] = Append("play", List("foo"))
我试图用它来处理这个问题,-Ytyper-debug
但它确实是硬核的东西,我不了解它的机制。
更新:我使用 Ordering 类型类复制了它,知道不满足什么隐式解析规则吗?
trait Req[I,O]
case class Insert[I : Ordering, O : Ordering](arg: I) extends Req[I,O]
def execute[I,O](req: Req[I,O]): O = null.asInstanceOf[O]
def main() = {
val result: Int = execute(Insert("test"))
}
error: type mismatch;
found : String
required: Int
val result: Int = execute(Insert("test"))