How do you declare a type alias in a scala constructor?

Question

If I have a class that takes a tuple in its constructor among other values such as:

class Foo(a: Int, b: String, c: (Int, String))

How do I use an abstract type to give the tuple a more descriptive name in a lightweight fashion (without wrapping it in a new class):

class Foo(a: Int, b: String, c: Dave)

I'm not sure how to bring a type alias in scope (or if this is the best thing to do):

type Dave = (Int, String)

Is there a convention for where to define types in this way (or should I be defining case classes and wrapping everything...)?

I appreciate that it does not make sense in a lot of situations but if I'm really only looking for a more descriptive name is it possible?

Thanks!

see [this](http://stackoverflow.com/questions/3400734/package-objects) — Jamil, Mar 21 '12 at 11:11
'type aliases' is a far better term than I managed to come up with... — Matthew Pickering, Mar 21 '12 at 11:24

score 10 · Accepted Answer · answered Mar 21 '12 at 12:29

10

You could use a type alias like:

scala> type MyTuple = Tuple2[Int,String]
defined type alias MyTuple

scala> val x = new MyTuple(1, "one")
x: (Int, String) = (1,one)

answered Mar 21 '12 at 12:29

jxstanford

4

Yes it was the scoping of the type alias that I was having trouble with, Jamil pointed me to package objects in the comment above. – Matthew Pickering Mar 21 '12 at 14:01
@MatthewPickering please close question – 0__ Dec 10 '15 at 10:51

1 Answers1