Scala operator overloading with multiple parameters

Question

In short: I try to write something like A <N B for a DSL in Scala, for an integer N and A,B of Type T. Is there a nice possibility to do so?

Longer: I try to write a DSL for TGrep2 in Scala. I'm currently interested to write

A <N B B is the Nth child of A (the rst child is <1).

in a nice way and as close as possible to the original definition in Scala. Is there a way to overload the < Operator that it can take a N and a B as a argument.

What I tried: I tried two different possibilities which did not make me very happy:

scala> val N = 10 N: Int = 10 scala> case class T(n:String) {def <(i:Int,j:T) = println("huray!")} defined class T scala> T("foo").<(N,T("bar")) huray!

and

scala> case class T(n:String) {def <(i:Int) = new {def apply(j:T) = println("huray!")}} defined class T scala> (T("foo")<N)(T("bar")) warning: there were 1 feature warnings; re-run with -feature for details huray!

Id suggest you use something like nth instead of the < symbol which makes the semantics clear. A nth N is B would make a lot of sense to me at least. — korefn
– korefn, Commented Apr 2, 2013 at 7:56
Scala's "operator" syntax sugar allows foo.bar(baz) to become foo bar baz, but it doesn't apply to functions with multiple arguments. You'll want to re-think your DSL so that your statement is more like foo.bar(baz).bang(biff) (A nth N is B) — Dylan
– Dylan, Commented Apr 2, 2013 at 11:20

korefn · Accepted Answer · 2013-04-02 08:24:34Z

Id suggest you use something like nth instead of the < symbol which makes the semantics clear. A nth N is B would make a lot of sense to me at least. It would translate to something like

case class T (label:String){ def is(j:T) = { label equals j.label } } case class J(i:List[T]){ def nth(index:Int) :T = { i(index) } }

You can easily do:

val t = T("Mice") val t1 = T("Rats") val j = J(List(t1,t)) j nth 1 is t //res = true

Mortimer · Accepted Answer · 2013-04-02 14:36:05Z

The problem is that apply doesn't work as a postfix operator, so you can't write it without the parantheses, you could write this:

 case class T(n: String) { def <(in: (Int, T)) = { in match { case (i, t) => println(s"${t.n} is the ${i} child of ${n}") } } } implicit class Param(lower: Int) { def apply(t: T) = (lower, t) }

but then,

T("foo") < 10 T("bar")

would still fail, but you could work it out with:

T("foo") < 10 (T("bar"))

there isn't a good way of doing what you want without adding parenthesis somewhere.

I think that you might want to go for a combinational parser instead if you really want to stick with this syntax. Or as @korefn proposed, you break the compatibility and do it with new operators.

Collectives™ on Stack Overflow

Scala operator overloading with multiple parameters

2 Answers 2

Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Related