在List的元素上定义约束函数？

Question

How can I define a function with the following signature,

f :: [Int???] -> [Int]
f xs = _ -- what I do with xs doesn't matter for the question

where a is a List of Int 's

such that the first argument's inputs, ie list elements, must be >= 0 , but <= 5 at compile-time?

In other words,

f [6] would fail to compile.

Answer 1

How about:

f :: [Int] -> [Int]
f = filter (\x -> x >= 0 && x <= 5)

Or do you want to enforce the bounds on the type (dependent types)?

If you want to restrict the range of the Int that is allowed you are probably better of using a smart constructor . Have a look here . The idea is that you create your own datatype and your own custom constructor:

newtype Range0_5 = Range0_5 { unRange :: Int }

makeRange0_5 :: Int -> Maybe Range0_5
makeRange0_5 x
  | x >= 0 && x <= 5 = Just $ Range0_5 x
  | otherwise        = Nothing

If you make a smart constructor , it is important to not expose it to the user of the module. This can be done by simply not exporting the Range0_5 constructor.

However this is not a compile time check. Other languages than Haskell might be more appropriate if you really need such a feature.

Since the range is fairly small, you could also make a sum type to represent it:

data Range0_5 = Int0 | Int1 | Int2 | Int3 | Int4 | Int5

Answer 2

If the signature is

f :: [Int] -> [Int]

(which was the original form of the question), then it is impossible to enforce your constraint at compile time. This follows from the standard diagonalization argument of the Halting problem .

Suppose the compiler could detect that

f[g x]

should not compile. By incorporating the source code of the compiler into g , it could choose the opposite of the compiler's decision.

---

Following your comment on Liquid Haskell (which seems like a very interesting project), note the following:

{-@ type Even = {v:Int | v mod 2 = 0} @-}

{-@ foo :: n:Even -> {v:Bool | (v <=> (n mod 2 == 0))} @-}   
foo   :: Int -> Bool 
foo n = if n^2 - 1 == (n + 1) * (n - 1) then True else foo (n - 1)

LiquidHaskell claims this function is unsafe, because, potentially foo n calls foo (n - 1) . Note, however, that this will never happen: it will only be called if the relationship n ² - 1 ≠ (n + 1) (n - 1) , which can never happen.

Again, this is not a criticism of the quality of LiquidHaskell, but rather just pointing out that it, too, cannot solve Halting Problem like issues.