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问题:
I have a Set in Scala (I can choose any implementation as I am creating the Set. The Java library I am using is expecting a java.util.Set[String].
Is the following the correct way to do this in Scala (using scala.collection.jcl.HashSet#underlying):
import com.javalibrary.Animals var classes = new scala.collection.jcl.HashSet[String] classes += "Amphibian" classes += "Reptile" Animals.find(classes.underlying)
It seems to be working, but since I am very new to Scala I want to know if this is the preferred way (any other way I try I am getting a type-mismatch error):
error: type mismatch; found : scala.collection.jcl.HashSet[String] required: java.util.Set[_]
回答1:
If you were asking about Scala 2.8, Java collections interoperability is supplied by scala.collection.JavaConversions. In this case, you want JavaConversions.asSet(...) (there's one for each direction, Java -> Scala and Scala -> Java).
For Scala 2.7, each scala.collection.jcl class that wraps a Java collection has an underlying property which provides the wrapped Java collection instance.
回答2:
Since Scala 2.12.0 scala.collection.JavaConversions is deprecated:
Therefore, this API has been deprecated and JavaConverters should be used instead. JavaConverters provides the same conversions, but through extension methods.
And since Scala 2.8.1 you can use scala.collection.JavaConverters for this purpose:
import scala.collection.JavaConverters._ val javaSet = new java.util.HashSet[String]() val scalaSet = javaSet.asScala val javaSetAgain = scalaSet.asJava
回答3:
回答4:
In Scala 2.12 it is possible to use : scala.collection.JavaConverters.setAsJavaSet(scalaSetInstance)
