The following Scala code works fine in Scala 2.9, but it generates compiler error in Scala 2.10 Milestone. Can anybody give me a hint how to create an ArrayTag:
type Lit = Array[Int]
var list = List[Lit].empty
list ::= Array(1,2,3)
list ::= Array(4,5)
val arr = list.toArray
Error message in Scala 2.10 is "No ArrayTag available for Lit".
Also, should I always create a new ArrayTag for new types?
Thank you.
That's a bug, I've lodged it: https://issues.scala-lang.org/browse/SI-5769
Related
Wrong syntax problem in recursively deleting scala files
Files.walk(path, FileVisitOption.FOLLOW_LINKS)
.sorted(Comparator.reverseOrder())
.forEach(Files.deleteIfExists)
The issue is that you're trying to pass a scala-style function to a method expecting a java-8-style function. There's a couple libraries out there that can do the conversion, or you could write it yourself (it's not complicated), or probably the simplest is to just convert the java collection to a scala collection that has a foreach method expecting a scala-style function as an argument:
import scala.collection.JavaConverters._
Files.walk(path, FileVisitOption.FOLLOW_LINKS)
.sorted(Comparator.reverseOrder())
.iterator().asScala
.foreach(Files.deleteIfExists)
In Scala 2.12 I expect this should work:
...forEach(Files.deleteIfExists(_: Path))
The reason you need to specify argument type is because expected type is Consumer[_ >: Path], not Consumer[Path] as it would be in Scala.
If it doesn't work (can't test at the moment), try
val deleteIfExists: Consumer[Path] = Files.deleteIfExists(_)
...forEach(deleteIfExists)
Before Scala 2.12, Joe K's answer is the correct one.
I'm having trouble iterating over a ImmutableList from Guava in Scala. The only reason I'm working with them is because I'm working with code written in Java that uses the Guava API. However, the compiler doesn't seem to like it. Here's my code:
for (blockData: IBlockData <- block.P.a) {
*insert actions here*
}
The compiler errors with this:
Error:(24, 43) value filter is not a member of com.google.common.collect.ImmutableList[net.minecraft.server.v1_8_R3.IBlockData]
for (blockData: IBlockData <- block.P.a) {
^
Any help is greatly appreciated.
Thanks!
To use Scalas for expression the object has to implement flatMap and filter which guava collections do not. Scala comes bundled with Java converters, this should work:
import collection.JavaConverters._
for (blockData: IBlockData <- block.P.a.asScala) {
*insert actions here*
}
I'm trying in Scala to get a list from a config file like something.conf with TypeSafe.
In something.conf I set the parameter:
mylist=["AA","BB"]
and in my Scala code I do:
val myList = modifyConfig.getStringList("mylist")
Simple configuration parameters works fine but could somebody give me an example of how to extract a list?
As #ghik notes, the Typesafe Config library is Java based, so you get a java.util.List[String] instead of a scala.List[String]. So either you make a conversion to a scala.List:
import collection.JavaConversions._
val myList = modifyConfig.getStringList("mylist").toList
Or (probably less awkward) you look for a Scala library. The tools wiki links at least to these maintained libraries:
Configrity
Bee Config
(Disclaimer: I don't use these, so you will have to check that they support your types and format)
For the record, since Scala 2.12 JavaConversions are deprecated so you can:
import collection.JavaConverters._
val myList: List[String] = modifyConfig.getStringList("mylist").asScala.toList
You can try my scala wrapper https://github.com/andr83/scalaconfig - it supports reading native scala types directly from config object. In your case it will look:
val myList = modifyConfig.as[List[String]]("mylist")
Starting Scala 2.13, the standard library provides Java to Scala implicit list conversions via scala.jdk.CollectionConverters:
import scala.jdk.CollectionConverters._
val myList: List[String] = conf.getStringList("mylist").asScala.toList
This replaces deprecated packages scala.collection.JavaConverters/JavaConversions.
This question already has answers here:
How do I get the Scala version from within Scala itself?
(4 answers)
Closed 1 year ago.
When trying to run some code in online interpreters or with IRC bots, I always wonder which version of Scala they support.
Is there a way to retrieve the version of Scala from within the interpreter?
For Scala 2, use scala.util.Properties.versionNumberString (or versionString):
scala> scala.util.Properties.versionString
val res0: String = version 2.13.6
scala> scala.util.Properties.versionNumberString
val res1: String = 2.13.6
For Scala 3, if you do the same thing, you may be surprised by the answer:
% scala3 -version
Scala code runner version 3.0.1 -- Copyright 2002-2021, LAMP/EPFL
% scala3
scala> scala.util.Properties.versionNumberString
val res0: String = 2.13.6
That's because Scala 3.0.x uses the Scala 2 standard library as-is, to aid migration, and makes only a small number of additions. (Eventually the standard libraries will no longer remain synchronized like this.)
Here's how to get the Scala 3 compiler version:
scala> dotty.tools.dotc.config.Properties.simpleVersionString
val res0: String = 3.0.1
This only works if the scala3-compiler JAR is on your classpath. (In the standard Scala 3 REPL, it is; in some other environments, it might not be.)
If the compiler isn't on your classpath and you want the full Scala 3 version string, see Dmitrii's answer.
If the compiler isn't on your classpath but you just want to find out at runtime whether you're on Scala 2 or 3, well... perhaps there's a cleaner/better way, you tell me, but one way that works is:
util.Try(Class.forName("scala.CanEqual")).isSuccess
Here, the choice of scala.CanEqual is arbitrary, it could be any of the small number of classes that are in scala3-library but not scala-library.
But if you are tempted to go that route, you might instead consider including version-specific source in your project, or passing the Scala version via sbt-buildinfo.
scala> scala.util.Properties.versionMsg
res: String = Scala library version 2.9.0.1 -- Copyright 2002-2011, LAMP/EPFL
Looks of course like the library version and not like the language version, but I think currently there won’t be a real difference in practice.
If you need just the version number without the "version" keyword you can use versionNumberString function.
scala> scala.util.Properties.versionNumberString
res1: String = 2.12.3
If you want to get the exact Scala 3 version, you can read it from the Manifest file
.../scala3-library_3-3.0.1.jar!/META-INF/MANIFEST.MF
import java.io.FileInputStream
import java.util.jar.JarInputStream
val scala3LibJar = classOf[CanEqual[_, _]].getProtectionDomain.getCodeSource.getLocation.toURI.getPath
val manifest = new JarInputStream(new FileInputStream(scala3LibJar)).getManifest
manifest.getMainAttributes.getValue("Implementation-Version")
Example in Scastie:
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[_]
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.
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
For 2.7.x I highly recommend using: http://github.com/jorgeortiz85/scala-javautils
Note that starting Scala 2.13, package scala.jdk.CollectionConverters replaces deprecated packages scala.collection.JavaConverters/JavaConversions._:
import scala.jdk.CollectionConverters._
// val scalaSet: Set[String] = Set("a", "b")
val javaSet = scalaSet.asJava
// javaSet: java.util.Set[String] = [a, b]
javaSet.asScala
// scala.collection.mutable.Set[String] = Set(a, b)
In Scala 2.12 it is possible to use : scala.collection.JavaConverters.setAsJavaSet(scalaSetInstance)