scala store method in set with possibility of removal - scala

Question

Here's some test code to demonstrate the problem:
import scala.collection.mutable.HashSet
import scala.collection.mutable.Set
object Test extends App
{
val set: Set[Int => Boolean] = new HashSet()
def test(i: Int): Boolean = {
false
}
Console.println(set.add(test)) // true
Console.println(set.remove(test)) // false
Console.println(set.add(test _)) // true
Console.println(set.remove(test _)) // false
}
What options are there for making this work?
Edit: I also want this to support lambdas.

Answer 1

Every time you perform "add" or "remove" operation Scala compiler creates new AbstractFunction1.mcZI.sp object instance:
Console..MODULE$.println(BoxesRunTime.boxToBoolean(set().add(
new AbstractFunction1.mcZI.sp() {
public final boolean apply(int i) { return apply$mcZI$sp(i); }
public boolean apply$mcZI$sp(int i) { return Test..MODULE$.test(i); }
})));
As I understand this object instance doesn't contain information about enclosing object and 'test' method which can be used for 'hashCode' and 'equals' methods.
I think as a workaround you may consider this:
import scala.collection.mutable.HashSet
import scala.collection.mutable.Set
object Test extends App
{
val set: Set[Int => Boolean] = new HashSet()
def test(i: Int): Boolean = {
false
}
val testFunc: Function1[Int, Boolean] = test
Console.println(set.add(testFunc)) // true
Console.println(set.remove(testFunc)) // true
val testPartFunc = test _
Console.println(set.add(testPartFunc)) // true
Console.println(set.remove(testPartFunc)) // true
}

Answer 2

If instead of passing methods you pass function it works:
val testFun: Int => Boolean = i => i > 10
set.add(testFun) // true
set.remove(testFun) // true
it will also work for anonymous functions:
set.add { _ => false }
but then I don't know how you would like to remove such function from the set.