Scalacheck: Generate list corresponding to list of generators

Question

I want to generate a list of integers corresponding to a list of generators in ScalaCheck.

 import org.scalacheck._ import Arbitrary.arbitrary val smallInt = Gen.choose(0,10) val bigInt = Gen.choose(1000, 1000000) val zeroOrOneInt = Gen.choose(0, 1) val smallEvenInt = smallInt suchThat (_ % 2 == 0) val gens = List(smallInt, bigInt, zeroOrOneInt, smallEvenInt) //val listGen: Gen[Int] = ?? //println(listGen.sample) //should print something like List(2, 2000, 0, 6)

For the given gens, I would like to create a generator listGen whose valid sample can be List(2, 2000, 0, 6). Here is my first attempt using tuples.

 val gensTuple = (smallInt, bigInt, zeroOrOneInt, smallEvenInt) val tupleGen = for { a <- gensTuple._1 b <- gensTuple._2 c <- gensTuple._3 d <- gensTuple._4 } yield (a, b, c, d) println(tupleGen.sample) // prints Some((1,318091,0,6))

This works, but I don't want to use tuples since the list of generators(gens) is created dynamically and the size of the list is not fixed. Is there a way to do it with Lists?

I want the use the generator of the list(listGen) in scalacheck forAll property checking.

This looks like a toy problem but this is the best I could do to create a standalone snippet reproducing the actual issue I am facing.

Eric · Accepted Answer · 2012-11-25 10:56:37Z

14

How about using the Gen.sequence method? It transforms an Iterable[Gen[T]] into a Gen[C[T]], where C can be List:

 def sequence[C[_],T](gs: Iterable[Gen[T]])(implicit b: Buildable[T,C]): Gen[C[T]] = ...

answered Nov 25, 2012 at 10:56

Eric

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1 Comment

dips Over a year ago

Works fine! I had to explicitly add the implicit argument because of ambiguity with another method. val listGen = Gen.sequence(gens)(util.Buildable.buildableList)

Joseph Lust · Accepted Answer · 2017-01-05 04:09:02Z

Just use Gen.sequence, but be careful as it will try to return a java.util.ArrayList[T] if you don't fully parameterize it (bug).

Full working example:

def genIntList(): Gen[List[Int]] = { val gens = List(Gen.chooseNum(1, 2), Gen.chooseNum(3, 4)) Gen.sequence[List[Int], Int](gens) } println(genIntList.sample.get) // prints: List(1,4)

Plasty Grove · Accepted Answer · 2012-11-25 16:27:51Z

1

EDIT: Please disregard, this doesn't answer the asker's question

I can't comment on posts yet, so I'll have to venture a guess here. I presume the function 'sample' applies to the generators

Any reason why you can't do:

gens map (t=>t.sample)

edited Nov 25, 2012 at 16:27

answered Nov 25, 2012 at 10:16

Plasty Grove

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2 Comments

dips Over a year ago

I don't want create samples manually. It will be done by scalacheck when I use the generator as forAll(listGen){ list => <list property> }

Plasty Grove Over a year ago

Ah sorry about that, didn't realise that scalacheck creates the samples

Ptharien's Flame · Accepted Answer · 2012-12-20 20:11:20Z

For a more theoretical answer: the method you want is traverse, which is equivalent to sequence compose map although it might be more efficient. It is of the general form:

def traverse[C[_]: Traverse, F[_]: Applicative, A, B](f: A => F[B], t: C[A]): F[C[B]]

It behaves like map but allows you to carry around some extra Applicative structure during the traversal, sequencing it along the way.

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