Can we use the singleton .type as a type parameter?

Question

I was cobbling together an answer to this question: Scala mixin to class instance, where I showed a way of "mixing-in" another trait or class instance to an existing instance: ``` case class Person(name: String) val dave = Person("Dave") val joe = Person("Joe") trait Dog { val dogName: String } val spot = new Dog { val dogName = "Spot" } implicit def daveHasDog(p: dave.type) = spot dave.dogName //"Spot" joe.dogName //error: value dogName is not a member of Person ``` So after the local implicit def, `dave` can effectively be used as a `Person with Dog`. My question is, if we wanted to define a method that takes a `Person` instance only where the `Person` has a `Dog`, how do we do it? I can define a method such as ``` def showDog(pd: Person with Dog) = pd.name + " shows " + pd.dogName ``` however this is no good for `dave` since he is still just a `Person`, despite his implicit transformation abilities. I tried defining ``` trait Dog [T] { val dogName: String } val spot = new Dog [dave…

Accepted Answer

If you compile with `-Ydependent-method-types`, your original code will work with this definition of `showDog`:

```
scala> def showDog(p: Person)(implicit ev: p.type => Dog) = p.name + " shows " + p.dogName
showDog: (p: Person)(implicit ev: p.type => Dog)java.lang.String

scala> showDog(dave)
res1: java.lang.String = Dave shows Spot
```

Can we use the singleton .type as a type parameter?

Solution

Problem