leaving out some implicit parameters

Question

Is it possible to leave out some implicit parameters but not all of them? I tried with named parameters:

def foo(implicit a: Int, b: String) {
  if (a > 0         


        

Answer 1

Not sure what you're trying to achieve, but something like this could do:

def foo(implicit a: Int, b: String): Unit = {
  def helper(a: Int)(implicit b: String): Unit =
    if (a > 0) {
      println(b)
      helper(a - 1)
    }
  helper(a)
}

Answer 2

It is not possible to leave out some implicit parameters. So, in your example

def foo(implicit a: Int, b: String): Unit = ???

It is not possible to only specify a. However, you can specify the default value of the implicit parameter, for example

def foo(implicit a: Int, b: String = "---"): Unit = ???

Where if b is not implicitly available, "---" will be used.

Remember that the implicit keyword marks the parameter list as implicit, not that one parameter as implicit.