Reputation: 12104
What would be the best and/or easiest way to extract a value that I've saved in a case class?
take for example the following code:
abstract class Something
case class Foo(input: Int) extends Something
case class Bar(input: Double) extends Something
def someMethod(a: Something, b: Something) {
// code that extracts values goes here
}
someMethod(Foo(10), Foo(20))
someMethod(Bar(2.1), Bar(21.2))
how would I then go about getting the integer or the double itself out of a
and b
when I call the method like I did under its definition?
Note that both the parameters are used in the same equation
Upvotes: 5
Views: 13625
Reputation: 1544
The other answers have covered the basic scenario. There are useful variations to consider.
Constructor Pattern
As already answered there is:
value match {
case Foo(x) => x
...
}
Deep Matching
The constructor pattern also supports deep matching. For example, extract x within Bar within Foo which is 3 levels deep:
value match {
case Foo(y, Bar(x)) => x
...
}
Variable Binding
If the value you want to extract is an actual case class inside another case class you can use variable binding. E.g. to extract the whole Bar(x) into b:
value match {
case Foo(y, b @ Bar(x)) => b
...
}
Programming in Scala by M. Odersky, Spoon and Venners has a great chapter on case classes and pattern matching which covers many other scenarios. Pattern matching is such a rich part of the language it would be a worthwhile investment.
Upvotes: 0
Reputation: 7373
In addition to the direct solution of pattern matching in your method, I'll try to show a somewhat more convoluted, general and functional approach to this kind of situations. Still pattern matching is the most direct and simple answer!
If you can explicitly "certify" in your interface the input
accessor, you can generalize how you work with the Something
class.
In code this translates to
trait Something[T] {
def input: T
}
case class Foo(input: Int) extends Something[Int]
case class Bar(input: Double) extends Something[Double]
from here you can define how to "lift" any function you like to one that works over Something
s
Let's say you have methods that takes two inputs (e.g. Ints
or Doubles
) and you want to operate on such inputs within one of your case classes (i.e. Foo
, Bar
)
//this function lift your specific input method to one that takes Somethings
def liftSomething2[T, R](f: (T, T) => R): (Something[T], Something[T]) => R =
(a, b) => f(a.input, b.input)
Let's examine this a bit: it takes a function
(T, T) => R
of 2 arguments of type T
and a result R
and transforms it in a
(Something[T], Something[T]) => R
which takes Something
s as arguments.
Examples
//lifts a function that sums ints
scala> val sumInts = liftSomething2[Int, Int](_ + _)
sumInts: (Something[Int], Something[Int]) => Int = <function2>
//lifts a function that multiplies ints
scala> val multInts = liftSomething2[Int, Int](_ * _)
multInts: (Something[Int], Something[Int]) => Int = <function2>
//lifts a function that divides doubles
scala> val divDbl = liftSomething2[Double, Double](_ / _)
divDbl: (Something[Double], Something[Double]) => Double = <function2>
//Now some test
scala> sumInts(Foo(1), Foo(2))
res2: Int = 3
scala> multInts(Foo(4), Foo(-3))
res3: Int = -12
scala> divDbl(Bar(20.0), Bar(3.0))
res4: Double = 6.666666666666667
//You can even complicate things a bit
scala> val stringApp = liftSomething2[Int, String](_.toString + _)
stringApp: (Something[Int], Something[Int]) => String = <function2>
scala> stringApp(Foo(1), Foo(2))
res5: String = 12
All the above examples lift functions of type (T,T) => R
but the "lifting" can be made for all and any argument you need
//This takes three args of different types and returns another type
// the logic doesn't change
def liftSomething3[A,B,C,R](f: (A,B,C) => R): (Something[A], Something[B], Something[C]) => R =
(a,b,c) => f(a.input, b.input, c.input)
//sums to ints and divides by a double
scala> val sumDiv = liftSomething3[Int,Int,Double,Double]((i,j,d) => (i + j) / d)
sumDiv: (Something[Int], Something[Int], Something[Double]) => Double = <function3>
scala> sumDiv(Foo(5), Foo(30), Bar(4.2))
res7: Double = 8.333333333333332
All we've seen so far should be somewhat related to category theory concepts like Applicative Functors and Comonads, but I'm no expert so I encourage you to search for yourself if you feel this sort of abstractions are useful and interesting.
Upvotes: 3
Reputation: 21547
In case classes constructor arguments are vals, so just call:
a.input
b.input
You can also use extractor with the help of unapply
method:
val Foo(val1) = a
val Bar(val2) = b
and then use val1
and val2
Update
Then you should use pattern matching on your value:
value match {
case Foo(val1) => val1
case Bar(val1) => val1
}
It works just like val Foo(val1) = a
, with using generated unapply
method (extractor) in your class, and it is also an expression, so you van assign the result to the variable
If you have multiple arguments just change PatMat construct according to the number of your parameters, in your case:
someMethod(a: Something, b: Something) = (a, b) match {
case (Foo(v1), Foo(v2)) => (v1, v2) // or any other logic with values
case (Foo(v1), Bar(v2)) => // logic for this case
... // logic for other cases
}
The more parameters the more cases you should provide, but you case blank cases if you don't need them
someMethod(a: Something, b: Something) = (a, b) match {
case (Foo(v1), Foo(v2)) => (v1, v2) // or any other logic with values
case _ =>
}
in this case all other cases will be ignored, not the best choice, cause the result type will be incorrect. And you also can black values
someMethod(a: Something, b: Something) = (a, b) match {
case (Foo(v1), _) => v1 // in such case you can work only with v1
... // logic for other cases
}
Upvotes: 10
Reputation: 35443
An alternative to pattern matching could be do redefine your classes like this:
trait Something[T]{
def input:T
}
case class Foo(input: Int) extends Something[Int]
case class Bar(input: Double) extends Something[Double]
Then, any instance of Something
will expose the input
property. The only potential downside is that it will be of a generic type when you access it.
Upvotes: 4
Reputation: 43309
In your example both the a
and b
have specific types: Foo
and Bar
respectively. That's why you can simply access their fields like so:
scala> a.input
res4: Int = 10
scala> b.input
res5: Double = 25.1
If however your value has type Something
, then you'll need to pattern-match:
val input = somethingOfTypeSomething match {
case Foo(input) => input
case Bar(input) => input
}
Upvotes: 1