scala中的函数如何匹配？

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?

Answer 1

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"