What's the difference between raw string interpolation and triple quotes in scala

Scala has triple quoted strings """String\nString""" to use special characters in the string without escaping. Scala 2.10 also added raw"String\nString" for the same purpose.

Is there any difference in how raw"" and """""" work? Can they produce different output for the same string?

Looking at the source for the default interpolators (found here: https://github.com/scala/scala/blob/2.11.x/src/library/scala/StringContext.scala) it looks like the "raw" interpolator calls the identity function on each letter, so what you put in is what you get out. The biggest difference that you will find is that if you are providing a string literal in your source that includes the quote character, the raw interpolator still won't work. i.e. you can't say

raw"this whole "thing" should be one string object"

but you can say

"""this whole "thing" should be one string object"""

So you might be wondering "Why would I ever bother using the raw interpolator then?" and the answer is that the raw interpolator still performs variable substitution. So

val helloVar = "hello"
val helloWorldString = raw"""$helloVar, "World"!\n"""

Will give you the string "hello, "World"!\n" with the \n not being converted to a newline, and the quotes around the word world.

It is surprising that using the s-interpolator turns escapes back on, even when using triple quotes:

scala> "hi\nthere."
res5: String =
hi
there.

scala> """hi\nthere."""
res6: String = hi\nthere.

scala> s"""hi\nthere."""
res7: String =
hi
there.

The s-interpolator doesn't know that it's processing string parts that were originally triple-quoted. Hence:

scala> raw"""hi\nthere."""
res8: String = hi\nthere.

This matters when you're using backslashes in other ways, such as regexes:

scala> val n = """\d"""
n: String = \d

scala> s"$n".r
res9: scala.util.matching.Regex = \d

scala> s"\d".r
scala.StringContext$InvalidEscapeException: invalid escape character at index 0 in "\d"
  at scala.StringContext$.loop$1(StringContext.scala:231)
  at scala.StringContext$.replace$1(StringContext.scala:241)
  at scala.StringContext$.treatEscapes0(StringContext.scala:245)
  at scala.StringContext$.treatEscapes(StringContext.scala:190)
  at scala.StringContext$$anonfun$s$1.apply(StringContext.scala:94)
  at scala.StringContext$$anonfun$s$1.apply(StringContext.scala:94)
  at scala.StringContext.standardInterpolator(StringContext.scala:124)
  at scala.StringContext.s(StringContext.scala:94)
  ... 33 elided

scala> s"""\d""".r
scala.StringContext$InvalidEscapeException: invalid escape character at index 0 in "\d"
  at scala.StringContext$.loop$1(StringContext.scala:231)
  at scala.StringContext$.replace$1(StringContext.scala:241)
  at scala.StringContext$.treatEscapes0(StringContext.scala:245)
  at scala.StringContext$.treatEscapes(StringContext.scala:190)
  at scala.StringContext$$anonfun$s$1.apply(StringContext.scala:94)
  at scala.StringContext$$anonfun$s$1.apply(StringContext.scala:94)
  at scala.StringContext.standardInterpolator(StringContext.scala:124)
  at scala.StringContext.s(StringContext.scala:94)
  ... 33 elided

scala> raw"""\d$n""".r
res12: scala.util.matching.Regex = \d\d