How function match in scala?

Question

case class Person(name:String,age:Int)

val p1 = Person("Maria",18)

def h(x:Person) = x match{
 case y if y.age >17 => "Adult"
 case z if z.age <=17 => "Younger"  
}

Why in the cases it refers to the age with y or z if the parameter containing the values is x?

Luis Miguel Mejía Suárez · Accepted Answer · 2020-02-19 14:33:18Z

6

It is because case y means pattern everything and name the variable y, so it is basically creating a new variable that refers to the same object.
However, this is a poor use of pattern matching.

A better alternative would be:

def h(person: Person): String = person match {
  case Person(_, age) if (age > 17) => "Adult"
  case _ => "Younger" // Checking the age here is redudant.
}

Or just use if / else:

def h(person: Person): String =
  if (person.age > 17) "Adult"
  else "Younger"

answered Feb 19, 2020 at 14:33

Luis Miguel Mejía Suárez

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2 Comments

Alexey Romanov Over a year ago

I'd disagree that the first alternative is better; changes to Person's definition are a lot more likely to break it.

Luis Miguel Mejía Suárez Over a year ago

@AlexeyRomanov Well, some may consider that a good thing, since changes on data model are usually meaningful. But, I do not want to discuss that. With "better" in this case, I meant about the use of pattern matching, because the original code wasn't really pattern matching anything. And for that reason, I also propose the if / else alternative which is shorter and more readable.

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