How to find the name of the enclosing source file in Scala 2.11

Question

At compile time, how to retrieve the name of the current source file (where the code is written) in scala 2.11?

Answer 1

In the REPL, the name is console, but this shows that a position knows its source.

scala> import scala.language.experimental.macros
import scala.language.experimental.macros

scala> import scala.reflect.macros.whitebox.Context
import scala.reflect.macros.whitebox.Context

scala> def f(c: Context): c.Tree = { import c._,universe._ ; Literal(Constant(c.enclosingPosition.source.file.name)) }
f: (c: scala.reflect.macros.whitebox.Context)c.Tree

scala> def g: String = macro f
defined term macro g: String

scala> g
res0: String = <console>

Answer 2

Here's a kinda hacky way that actually does the trick:

val srcFile = new Exception().getStackTrace.head.getFileName
println(srcFile)

Source: How do I get the current script or class name in Scala?

Answer 3

Or, you can use @li-haoyi's excellent sourcecode macro library

def foo(arg: String)(implicit file: sourcecode.File) = {
  println(s"this file is: '${file.getAbsolutePath}'")
}

foo("hello") // "this file is '/path/to/file'"

Plus, many other goodies like line number, variables names, etc.