FsCheck custom generator of lists satisfying a predicate

Question

For a given predicate pred : 'a list -> bool and generator gen : Gen<'a>, consider the following generator of well-formed lists that satisfy the predicate: let wellFormedList pred gen = Gen.ofList gen |> Gen.filter pred As mentioned in the FsCheck manual, there should be a high chance that predicate holds for a random list. Unfortunately, this assumption does not hold in my case. Thus, I need te define a custom generator for lists that satisfy the predicate. How can I define a custom generator that starts from the empty list and extends it with new random elements, until the list satisfies the predicate? I probably need to use the computation expression gen { } for generators, but I do not see how. PS: I am aware that, unlike the original implementation of wellFormedList, the distribution of such a custom generator is not uniform over all lists that satisfy the predicate. JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

Well if I'm understanding it right, I think this should do it: let wellFormedList (predicate:'a list -> bool) (myGen:Gen<'a>) = gen { let! initialList = Gen.listOf myGen let mutable myList = initialList while not <| predicate myList do let! newVal = myGen myList <- (newVal :: myList) return myList } or if you wanted a recursive function: let wellFormedList (predicate:'a list -> bool) (myGen:Gen<'a>) = gen { let rec addIfNecessary listSoFar = if predicate listSoFar then gen { return listSoFar } else gen { let! newVal = myGen return! addIfNecessary (newVal :: listSoFar) } let! initialList = Gen.listOf myGen return! addIfNecessary initialList } I haven't checked what kind of distributions that gives you though. And of course you risk an infinite loop (or stack overflow) if the list never converges on a form that passes the predicate.