Difference in singleton object with parentheses and without in pattern matching?

user_11934502 0
Viewed 46

On using even() is it so that it only matches the true condition from the unapply and on using even , how does this still prints "Is even"

object even{
  def unapply(arg: Int): Boolean = (arg%2 == 0)
}
object singleDigit{
  def unapply(arg: Int): Option[Boolean] = {
    if(arg < 10) Some(true)
    else None
  }
}
val n = 9
n match {
  case even() => println("Is even")
  case singleDigit => println("is single")
  case _ => println("No property")
}
//This gives "is single" 

n match {
  case even => println("Is even")
  case singleDigit => println("is single")
  case _ => println("No property")
}
//This gives "Is even"

How is even case satisfied here being n=9 as an odd number ? what is the difference between even and even() inside this pattern matching?

pattern-matching scala functional-programming
1 Answers

The case even => is actually a variable pattern and is not referring to even.unapply as hinted by compiler warning

Warning:(19, 9) patterns after a variable pattern cannot match (SLS 8.1.1)
If you intended to match against object even in class A$A3, you must use backticks, like: case `even` =>
  case even => println("Is even")
       ^

You just happened to have given it the same name, but you could write 

n match {
  case foo => println("Is even")
  ...
}

and it would still print out Is even because the variable pattern is in the first position.

user_52050220
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