Generating a lists of a specific length with Haskell's QuickCheck

Question

-- 3 (find k"th element of a list)
element_at xs x = xs !! x
prop_3a xs x = (x < length xs && x >= 0) ==> element_at xs (x::Int) == (xs !! x::Int)

When prop_3a is ran through QuickCheck, it gives up, because it won't generate long enough lists.

How can I write a generator that will generate lists with length longer than the random integer?

Answer 1

hammar's answer is perfectly adequate for the problem. But for the sake of answering the precise question asked, I couldn't help but investigate a bit. Let's use forAll.

prop_bang x = x >= 0 ==> forAll (listLongerThan x) $ \xs ->
  element_at xs x == xs !! x

So now we need a function, listLongerThan :: Int -> Gen [Int]. It takes a length, x, and produces a generator which will produce lists of length greater than x.

listLongerThan :: Int -> Gen [Int]
listLongerThan x = replicateM (x+1) arbitrary

It's rather straightforward: we simply take advantage of the Monad instance of Gen. If you run quickCheck prop_bang, you'll notice it starts taking quite a long time, because it begins testing absurdly long lists. Let's limit the length of the list, to make it go a bit faster. Also, right now listLongerThan only generates a list that is exactly x+1 long; let's mix that up a bit, again utilizing the Monad instance of Gen.

prop_bang =
  forAll smallNumber $ \x ->
  forAll (listLongerThan x) $ \xs ->
  element_at xs x == xs !! x

smallNumber :: Gen Int
smallNumber = fmap ((`mod` 100) . abs) arbitrary

listLongerThan :: Int -> Gen [Int]
listLongerThan x = do
  y <- fmap (+1) smallNumber -- y > 0
  replicateM (x+y) arbitrary

You can use sample smallNumber or sample (listLongerThan 3) in ghci to make sure it is generating the correct stuff.