Does Scala have record update syntax for making modified clones of immutable data structures?

Question

In Mercury I can use:

A = B^some_field := SomeValue

to bind A to a copy of B, except that some_field is SomeValue instead of whatever it was in B. I believe the Haskell equivalent is something like:

a = b { some_field = some_value }

Does Scala have something like this for "modifying" immutable values. The alternative seems to be to have a constructor that directly sets every field in the instance, which isn't always ideal (if there are invarients the constructor should be maintaining). Plus it would be really clunky and much more fragile if I had to explicitly pass every other value in the instance I want to have a modified copy of.

I couldn't find anything about this by googling, or in a brief survey of the language reference manual or "Scala By Example" (which I have read start-to-finish, but haven't absorbed all of yet, so it may well be in there).

I can see that this feature could have some weird interactions with Java-style access protection and subclasses though...

Answer 1

If you define your class as a case class, a convenient copy method is generated, and calling it you can specify with named parameters new values for certain fields.

scala> case class Sample(str: String, int: Int)
defined class Sample

scala> val s = Sample("text", 42)
s: Sample = Sample(text,42)

scala> val s2 = s.copy(str = "newText")
s2: Sample = Sample(newText,42)

It even works with polymorphic case classes:

scala> case class Sample[T](t: T, int: Int)
defined class Sample

scala> val s = Sample("text", 42)
s: Sample[java.lang.String] = Sample(text,42)

scala> val s2 = s.copy(t = List(1,2,3), 42)
s2: Sample[List[Int]] = Sample(List(1, 2, 3),42)

Note that s2 has a different type than s.