How to check constructor arguments and throw an ex

问题:

I would like to check constructor arguments and refuse to construct throwing IllegalArgumentException in case the arguments set is not valid (the values don't fit in expected constraints). How to code this in Scala?

回答1:

In Scala, the whole body of the class is your primary constructor, so you can add your validation logic there.

scala> class Foo(val i: Int) {
     |   if(i < 0) 
     |     throw new IllegalArgumentException("the number must be non-negative.")
     | }
defined class Foo

scala> new Foo(3)
res106: Foo = Foo@3bfdb2

scala> new Foo(-3)
java.lang.IllegalArgumentException: the number must be positive.

Scala provides a utility method require that lets you write the same thing more concisely as follows:

class Foo(val i: Int) {
  require(i >= 0, "the number must be non-negative.")
}

A better approach might be to provide a factory method that gives a scalaz.Validation[String, Foo] instead of throwing an exception. (Note: requires Scalaz)

scala> :paste
// Entering paste mode (ctrl-D to finish)

class Foo private(val i: Int)

object Foo {
  def apply(i: Int) = {
    if(i < 0)
      failure("number must be non-negative.")
    else
      success(new Foo(i))
  }
}

// Exiting paste mode, now interpreting.

defined class Foo
defined module Foo

scala> Foo(3)
res108: scalaz.Validation[java.lang.String,Foo] = Success(Foo@114b3d5)

scala> Foo(-3)
res109: scalaz.Validation[java.lang.String,Foo] = Failure(number must be non-negative.)

回答2:

scala> class Foo(arg: Int) {
     |   require (arg == 0)
     | }
defined class Foo

scala> new Foo(0)
res24: Foo = Foo@61ecb73c

scala> new Foo(1)
java.lang.IllegalArgumentException: requirement failed