Based on my knowledge so far you cannot define a type with default parameters. A type is a type.
What you can do is define a partially applied function.
Taking as an example the following function:
scala> def factorOf(x: Int, y: Int) = y % x == 0
factorOf: (x: Int, y: Int)Boolean
If you want a shortcut to the function without retaining any parameters, you can use the wildcard operator (_) assignment
scala> val f = factorOf _
f: (Int, Int) => Boolean = <function2>
scala> val x = f(7, 20)
x: Boolean = false
If you want to retain some of the parameters, you can partially apply the function by using the wildcard operator to take the place of one of the parameters. The wildcard operator here requires an explicit type, because it is used to generate a function value with a declared input type:
scala> val multipleOf3 = factorOf(3, _: Int)
multipleOf3: Int => Boolean = <function1>
scala> val y = multipleOf3(78)
y: Boolean = true
The new function value, multipleOf3, is a partially applied function, because it contains some but not all of the parameters for the factorOf() function.
A cleaner way to partially apply functions is to use functions with multiple parameter lists. This is a technique
known as currying the function:
scala> def factorOf(x: Int)(y: Int) = y % x == 0
factorOf: (x: Int)(y: Int)Boolean
scala> val isEven = factorOf(2) _
isEven: Int => Boolean = <function1>
scala> val z = isEven(32)
z: Boolean = true