如何在scala中编写一个以函数为参数的类

Question

I have a function, how to make this function as a parameter for ServicesImpl class?

def blacklist: List[String] = {
    ***
  }

class ServicesImpl(){}

Answer 1

This compiles.

class ServicesImpl(ss :List[String])

def blacklist: List[String] = List("")
new ServicesImpl(blacklist)

Answer 2

Maybe you want this

class ServicesImpl(foo: => List[String]) {}

def blacklist: List[String] = { }

val impl = new ServicesImpl(blacklist)