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Home/ Questions/Q 6081969
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T11:14:11+00:00 2026-05-23T11:14:11+00:00

As asked in this thread on the Scala mailing list, how can I create

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As asked in this thread on the Scala mailing list, how can I create an embedded Scala REPL that inherits the classpath of the parent program? Suppose the parent Scala program is launched using scala -cp <classpath> ...; can <classpath> be accessed as a string and used to initialize the embedded REPL? (The Java classpath, available via System.getProperty("java.class.path"), appears to differ from the Scala classpath.)

Alternatively, perhaps the embedded Scala REPL can inherit or construct its ClassLoader from the parent process (Michael Dürig’s ScalaDays 2010 talk might be relevant). Is this the recommended approach?

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  1. Editorial Team
    Editorial Team
    2026-05-23T11:14:12+00:00Added an answer on May 23, 2026 at 11:14 am

    I’m trying to do the same thing, and I just found a way my out by Googling:

    lazy val urls = java.lang.Thread.currentThread.getContextClassLoader match {
      case cl: java.net.URLClassLoader => cl.getURLs.toList
      case _ => error("classloader is not a URLClassLoader")
    }
    lazy val classpath = urls map {_.toString}
    

    The above code gets you the classpath in current context.

    settings.classpath.value = classpath.distinct.mkString(java.io.File.pathSeparator)
    

    Put that into your settings.classpath and you should be able to fire up dispatch or whatever library you need.

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