How to use a specific file name in a scala exec bash file?

Question

How would I write the following bash script replacing the actual code of the scala section (after the !#) line to be called from a .scala file and then call its main method?

#!/bin/sh
exec scala "$0" "$@"
!#
object HelloWorld {
  def main(args: Array[String]) {
    println("Hello, world! " + args.toList)
  }
}
HelloWorld.main(args)

keep in mind i do not have access to scalac

Maybe this is what you're actually trying to do: stackoverflow.com/questions/15077341/… — Kenster
– Kenster, Commented Jul 10, 2015 at 20:10

Eugeniu Rosca · Accepted Answer · 2015-07-10 19:28:15Z

1

I presume you mean dividing the file into two parts...

file.scala

object HelloWorld {
  def main(args: Array[String]) {
    println("Hello, world! " + args.toList)
  }
}
HelloWorld.main(args)

launch.sh

#!/bin/sh
exec scala "file.scala" "$@"

edited Jul 10, 2015 at 19:28

answered Jul 10, 2015 at 19:22

Eugeniu Rosca

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2 Comments

Zee Over a year ago

YES! this is exactly what I was looking for! out of curiosity what does the "$@" stand for?

Eugeniu Rosca Over a year ago

@Zee: Source $@ is bash internal/special variable which is expanded to all the parameters received (in this particular case by launch.sh). That means if you call ./launch.sh arg1 arg2, then "$@" will be expanded to arg1 arg2

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