Is there a way an implicit class can override a default implementation

Question

Say I define the following:

class A {
    def foo() = println("A::foo")
}

implicit class Wrapper(a: A) {
    def foo() = println("Wrapper::foo")
    def bar() = println("Wrapper::bar")
}

val a = new A
a.foo()
a.bar()

A::foo() is the method that is called. Is there any possible way an implicit class can override the default implementation in A?

Answer 1

If a member method exists then compiler would not search for implicit. If overriding with inheritance is not an option, you might consider "turning the tables" and create a delegate like so

class A {
  def foo() = println("A::foo")
  def zar() = println("A::zar")
}

class Wrapper(val a: A) {
  def foo() = println("Wrapper::foo")
  def bar() = println("Wrapper::bar")
}

implicit def unwrap(wrapped: Wrapper): A = wrapped.a

val a = new Wrapper(new A)
a.zar()   // A::zar
a.foo()   // Wrapper::foo
a.bar()   // Wrapper::bar