Question

I’d like to read a string as an instance of a case class. For example, if the function were named “read” it would let me do the following:

case class Person(name: String, age: Int)
val personString: String = "Person(Bob,42)"
val person: Person = read(personString)

This is the same behavior as the read typeclass in Haskell.

Answer 1

dflemstr answered more towards setting up the actual read method- I’ll answer more for the actual parsing method.

My approach has two objects that can be used in scala’s pattern matching blocks. AsInt lets you match against strings that represent Ints, and PersonString is the actual implementation for Person deserialization.

object AsInt {
  def unapply(s: String) = try{ Some(s.toInt) } catch {
    case e: NumberFormatException => None
  }
}

val PersonRegex = "Person\\((.*),(\\d+)\\)".r

object PersonString {
  def unapply(str: String): Option[Person] = str match {
    case PersonRegex(name, AsInt(age)) => Some(Person(name, age))
    case _ => None
  }
}

The magic is in the unapply method, which scala has syntax sugar for. So using the PersonString object, you could do

val person = PersonString.unapply("Person(Bob,42)")
//  person will be Some(Person("Bob", 42))

or you could use a pattern matching block to do stuff with the person:

"Person(Bob,42)" match {
  case PersonString(person) => println(person.name + " " + person.age)
  case _ => println("Didn't get a person")
}