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?

The compiler will never go to an implicit if it doesn't have to. It's always the solution of last resort. — jwvh, Jan 25 '20 at 16:04

score 4 · Accepted Answer · answered Jan 25 '20 at 16:34

4

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

answered Jan 25 '20 at 16:34

Mario Galic

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Can you plz explain what does "implicit def unwrap(wrapped: Wrapper): A = wrapped.a" do? – Salim Jan 25 '20 at 17:58
1

@Salim It expands `a.zar()` into `unwrap(a).zar()` because `Wrapper` does not have `zar` member but `A` does. This delegates all calls to `A` except for those methods provided in `Wrapper`, such as `foo`, which in effect enables us to "override" `A.foo`. – Mario Galic Jan 25 '20 at 18:11
Thanks Mario. Appreciate it. – Salim Jan 25 '20 at 19:32

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

1 Answers1