Scala: Is it possible to indicate a generic class which implements a certain method

Question

I thinks it's easier to explain it with a simple example. (help rephrasing the title is welcome ;-)

I'd like to implement a squared method and, using implicit def, automatically add it to any class that supports the *-operator.

With an Int it's very easy:

class EnhancedInt(x: Int) { def squared = x * x }

implicit def IntToEnchancedInt(x: Int) = new EnhancedInt(x)

But with Any or AnyVal I get the following error:

scala> class EnhanceAny(x: AnyVal) { def squared = x * x }
<console>:7: error: value * is not a member of AnyVal
       class EnhanceAny(x: AnyVal) { def squared = x * x }

I'd like to know how I could apply it to any numeric class, or, even better, to any class supporting the *-operator.

Answer 1

It's not possible to have solution that works on any type with a * method without writing a boilerplate conversion for each type you want to deal with. Essentially to do that you would need a recursive structural type, and Scala does not support those because of JVM type erasure. See this post for more details.

You can get fairly close to what you want using a type class along with the Numeric type class, (inspired by the answers to this and this question). This will work with most primitives:

//define the type class
trait Multipliable[X] { def *(x: X): X}

//define an implicit from A <% Numeric[A] -> Multipliable[Numeric[A]]
implicit def Numeric2Mult[A](a: A)(implicit num: Numeric[A]): Multipliable[A] = new Multipliable[A]{def *(b: A) = num.times(a, b)}

//now define your Enhanced class using the type class
class EnhancedMultipliable[T <% Multipliable[T]](x: T){ def squared = x * x}

//lastly define the conversion to the enhanced class
implicit def Mult2EnhancedMult[T <% Multipliable[T]](x: T) = new EnhancedMultipliable[T](x)

3.squared
//Int = 9

3.1415F.squared
//Float = 9.869022

123456789L.squared
//Long = 15241578750190521