I am using IntelliJ.
To compiled my .scala file, at command prompt, and it created Item.Class file.
In IntelliJ, I created scala script file and tried to Instantiate the Item object
Item("Jacket", 10)
It says "Cannot resolve Item". I know this is a basic question. What is the best way to reference a compiled class in script files?
case class Item (name:String, quantity:Int) extends AnyVal{
}
Related
i created a scala spark project on intellij and created the executable jar file. However when i open up the command prompt, go to the jar file location and enter
java -jar ScalaProject.jar
It throws an error saying:
Error: cound not find or load main class SparkProject
I have tried removing my main method and extending APP like so
object SparkProject extends APP {
}
but it still shows the errors. Any ideas??
I need to compile a scala code which calls a java code from it.
What I did:
1]I have a scala main file check.scala
package com.code
class check {
var Rectvalue = Array.ofDim[Int](5)
val str = Array.ofDim[String](1)
def nativeacces(arg: String, loop: Integer) {
val test = new testing()
test.process(arg, Rectvalue,str)
}
}
2.For creating instance val test = new testing() ,i added two .class(sample.class,testJNI.class) file from java source code inside the folder(package) com/code.
3.When I compile the scala code using
scalac check.scala
It generates the class file for the scala file.
What I have to do:
1.Instead of .class(sample.class,testJNI.class) file added inside the package ,i need to add jar file.
2.I tried, created jar file for the .class file and compile the Scala, it shows the error:
scala:6: error: not found: type testing
val test = new testing()
3.I need to link the .jar file and compile the scala main file
You can reference classes/directories/JARs via classpath option:
scalac -classpath your.jar check.scala
Related question: Adding .jar's to classpath (Scala).
If you want a proper build use SBT, put your JAR in lib directory in the root of project and it will figure out what to do for you. Here is Hello World of SBT.
New to Scala and having problems reading an XML file in a Scala worksheet. So far I have:
downloaded the Scala IDE (for Windows) and unzipped it to my C:\ drive
created a Scala project with the following file path: C:\eclipse\workspace\xml_data
created the xml file ...\xml_data\music.xml using the following data
created a package sample_data and create the following object (with file path: ...\xml_data\src\sample_data\SampleData.scala):
package sample_data
import scala.xml.XML
object SampleData {
val data = XML.loadFile("music.xml")
}
object PrintSampleData extends Application {
println(SampleData.data)
}
This runs OK, however, when I create the Scala worksheet test_sample_data.sc:
import sample_data.SampleData
object test {
println(SampleData.data)
}
I get a java.lang.ExceptionInInitializerError which includes: Caused by: java.io.FileNotFoundException: music.xml (The system cannot find the file specified).
The workspace is C:\eclipse\workspace. Any help or insight much appreciated. Cheers!
UPDATE:
Following aepurniet's advice, I ran new java.io.File(".").getAbsolutePath() and got the following respectively:
SampleData.scala: C:\eclipse\workspace\xml_data\.
test_sample_data.sc: C:\eclipse\.
So this is what is causing the problem. Does anyone know why this occurs? Absolute file paths resolve the problem. Is this the best solution?
Regarding what is causing different user directory between the scala class and worksheet:
You are likely hitting the Eclipse IDE issue listed here
https://github.com/scala-ide/scala-worksheet/issues/102
Jfyi, I used Intellij and the issue is not reproducible there.
Regarding using absolute paths:
Using absolute path works fine for quick testing, but would NOT be a good practice for the actual implementation. You can consider passing the path along with the filename as input to SampleData.
Some hack mentioned here to get the base path of the workspace from the scala worksheet: Configure working directory of Scala worksheet
If this is just for your testing, hacking the absolute path of workspace inside the worksheets might be the easiest for you.
SampleData.scala
package sample_data
import scala.xml.XML
object SampleData {
def data(filename: String) = XML.loadFile(filename)
}
object PrintSampleData extends Application {
println(SampleData.data(System.getProperty("user.dir") + "/music.xml")
}
Scala worksheet:
import sample_data.SampleData
object test {
val workDir = ... // Using the hack or hardcoding
println(SampleData.data(workDir + "/music.xml"))
}
I am learning Scala so bear with me if this is a stupid question.
I have this package and a class (teared it down to most simplistic version):
package Foo {
class Bar {}
}
then in main.scala file I have:
import Foo.Bar
object test {
def main() {
val b = new Bar()
}
}
Why am I getting this:
test.scala:1: error: Bar is not a member of Foo
It points at the import statement.
scalac is the scala compiler. Foo.bar needs to have been compiled for you to use it, so you can't just run your main.scala as a script.
The other mistake in your code is that the main method needs to be
def main(args: Array[String]) { ...
(or you could have test extends App instead and do away with the main method).
I can confirm if you put the two files above in an empty directory (with the correction on the main method signature) and run scalac * followed by scala test it runs correctly.
The most likely explanation is that you did not compile the first file, or you are doing something wrong when compiling. Let's say both files are in the current directory, then this should work:
scalac *.scala
It should generate some class files in the current directory, as well as a Bar.class file in the Foo directory, which it will create.
To quickly test a scala code in IntelliJ (with the Scala plugin), you can simply type Ctrl+Shift+F10:
Note that for testing a Scala class, you have other choices, also supported in IntelliJ:
JUnit or TestNG
ScalaTest
I'm trying to compile the program with 2 simplest classes:
class BaseClass
placed in BaseClass.scala and
class Test extends BaseClass
placed in Test.scala. Issuing command scalac Test.scala fails, cause BaseClass is not found.
I don't want to compile classes one by one or using scalac *.scala.
The same operation in java works: javac Test.java. Where am I wrong?
Let's see first what Java does:
dcs#dcs-132-CK-NF79:~/tmp$ ls *.java
BaseClass.java Test.java
dcs#dcs-132-CK-NF79:~/tmp$ ls *.class
ls: cannot access *.class: No such file or directory
dcs#dcs-132-CK-NF79:~/tmp$ javac -cp . Test.java
dcs#dcs-132-CK-NF79:~/tmp$ ls *.class
BaseClass.class Test.class
So, as you can see, Java actually compiles BaseClass automatically when you do that. Which begs the question: how can it do that? Can does it know what file to compile?
Well, when you write extends BaseClass in Java, you actually know a few things. You know the directory where these files are found, from the package name. It also knows BaseClass is either in the current file, or in a file called BaseClass.java. If you doubt either of these facts, try moving the file from directory or renaming it, and see if Java can compile it.
So, why can't Scala do the same? Because it assumes neither thing! Scala's files can be in any directory, irrespective of the package they declare. In fact, a single Scala file can even declare more than one package, which would make the directory rule impossible. Also, a Scala class can be in any file whatsoever, irrespective of its name.
So, while Java dictates to you what directory the file should be in and what the file is called, and then reaps the benefit by letting you omit filenames from the command line of javac, Scala let you organize your code in whatever way seems best to you, but requires you to tell it where that code is.
Take your pick.
You need to compile BaseClass.scala first:
$ scalac Test.scala
Test.scala:1: error: not found: type BaseClass
class Test extends BaseClass
^
one error found
$ scalac BaseClass.scala
$ scalac Test.scala
$
EDIT So, the question is now why you have to compile the files one by one? Well, because the Scala compiler just doesn't do this kind of dependency handling. Its authors probably expect you to use a build tool like sbt or Maven so that they don't have to bother.