Check file name correctness during compilation using Scala interpolation / macros.

FileNameImpl.scala

package utils

import scala.language.experimental.macros
import scala.reflect.macros.Context

object FileNameImpl {
  def fileImpl(c: Context)(): c.Expr[String] = {
    import c.universe._

    val Apply(_, List(Apply(_, List(Literal(Constant(fileName: String)))))) = c.prefix.tree
    fileName match {
      case validFileName() =>
        c.Expr[String](Literal(Constant(fileName)))
      case _ =>
        c.error(c.enclosingPosition,"string does not confirms to file name regex ^[a-zA-Z0-9._ -]+$")
        c.Expr[String](Literal(Constant("")))
    }
  }

  lazy val validFileName = """^[a-zA-Z0-9._ -]+$""".r
}

gistfile1.scala

// Now, instead of accidentally writing:

val myFile = new File("some_file_with)underscores")

// You can write:

import utils.Utils

scala> val myFile = new File(file"some_file_with)underscores") // You will get a compile-time error here

//<console>:11: error: string does not confirms to file name regex ^[a-zA-Z0-9._ -]+$
//       val myFile = new File(file"some_file_with)underscores")

//But this will work fine:

scala>   val myFile = new File(file"some_file_with_underscores")
myFile: java.io.File = some_file_with_underscores

Utils.scala

package utils

import scala.language.experimental.macros

object Utils {

  /**
   * This will allow you to write file"some_file_name" which will be evaluated at compile time!
   */
  implicit class FileNameContext(val sc: StringContext) extends AnyVal {
    def file():String = macro FileNameImpl.fileImpl
  }
}