[英]Treat scala function with default parameter type as if it did not have default parameters
免責聲明:我是Scala的新手。
我想傳遞帶有默認參數的函數,就像其類型沒有默認參數一樣
import scala.util.hashing.MurmurHash3
type Record = Map[String, String]
type Dataset = Seq[Record]
def dropDuplicates(dataset: Dataset, keyF: Record => Any = recordHash) : Dataset = {
// keep only distinct records as defined by the key function
// removed method body for simplicity
return dataset
}
def recordHash(record: Record, attributes: Option[Seq[String]] = None) : Int = {
val values : Seq[String] = attributes
.getOrElse(record.keys.toSeq.sorted)
.map(attr => record(attr))
return MurmurHash3.seqHash(values)
}
這是我在編譯時遇到的錯誤:
error: type mismatch;
[ant:scalac] found : (Record, Option[Seq[String]]) => Int
[ant:scalac] required: Record => Any
[ant:scalac] def dropDuplicates(dataset: Dataset, keyF: Record => Any = recordHash) : Dataset = {
直觀地,當未提供默認參數attributes
時,我認為recordHash
為Record => Int
類型。 有沒有一種方法可以將recordHash
視為Record => Int
類型?
我無法編譯您的代碼,因為我錯過了一些類型,但是我認為這可以工作。
def dropDuplicates(dataset: Dataset, keyF: Record => Any = recordHash(_)) : Dataset = {
// keep only distinct records as defined by the key function
}
之所以recordHash(_)
是因為recordHash(_)
等效於x => recordHash(x)
,這樣, x
(函數的輸入)就是Record
,它就是您想要的類型。
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