When do I need type annotations? - haskell

Question

Consider these functions
{-# LANGUAGE TypeFamilies #-}
tryMe :: Maybe Int -> Int -> Int
tryMe (Just a) b = a
tryMe Nothing b = b
class Test a where
type TT a
doIt :: TT a -> a -> a
instance Test Int where
type TT Int = Maybe Int
doIt (Just a) b = a
doIt (Nothing) b = b
This works
main = putStrLn $ show $ tryMe (Just 2) 25
This doesn't
main = putStrLn $ show $ doIt (Just 2) 25
{-
• Couldn't match expected type ‘TT a0’ with actual type ‘Maybe a1’
The type variables ‘a0’, ‘a1’ are ambiguous
-}
But then, if I specify the type for the second argument it does work
main = putStrLn $ show $ doIt (Just 2) 25::Int
The type signature for both functions seem to be the same. Why do I need to annotate the second parameter for the type class function? Also, if I annotate only the first parameter to Maybe Int it still doesn't work. Why?

Answer 1

When do I need to cast types in Haskell?
Only in very obscure, pseudo-dependently-typed settings where the compiler can't proove that two types are equal but you know they are; in this case you can unsafeCoerce them. (Which is like C++' reinterpret_cast, i.e. it completely circumvents the type system and just treats a memory location as if it contains the type you've told it. This is very unsafe indeed!)
However, that's not what you're talking about here at all. Adding a local signature like ::Int does not perform any cast, it merely adds a hint to the type checker. That such a hint is needed shouldn't be surprising: you didn't specify anywhere what a is supposed to be; show is polymorphic in its input and doIt polymorphic in its output. But the compiler must know what it is before it can resolve the associated TT; choosing the wrong a might lead to completely different behaviour from the intended.
The more surprising thing is, really, that sometimes you can omit such signatures. The reason this is possible is that Haskell, and more so GHCi, has defaulting rules. When you write e.g. show 3, you again have an ambiguous a type variable, but GHC recognises that the Num constraint can be “naturally” fulfilled by the Integer type, so it just takes that pick.
Defaulting rules are handy when quickly evaluating something at the REPL, but they are fiddly to rely on, hence I recommend you never do it in a proper program.
Now, that doesn't mean you should always add :: Int signatures to any subexpression. It does mean that, as a rule, you should aim for making function arguments always less polymorphic than the results. What I mean by that: any local type variables should, if possible, be deducable from the environment. Then it's sufficient to specify the type of the final end result.
Unfortunately, show violates that condition, because its argument is polymorphic with a variable a that doesn't appear in the result at all. So this is one of the functions where you don't get around having some signature.

Answer 2

All this discussion is fine, but it hasn't yet been stated explicitly that in Haskell numeric literals are polymorphic. You probably knew that, but may not have realized that it has bearing on this question. In the expression
doIt (Just 2) 25
25 does not have type Int, it has type Num a => a — that is, its type is just some numeric type, awaiting extra information to pin it down exactly. And what makes this tricky is that the specific choice might affect the type of the first argument. Thus amalloy's comment
GHC is worried that someone might define an instance Test Integer, in which case the choice of instance will be ambiguous.
When you give that information — which can come from either the argument or the result type (because of the a -> a part of doIt's signature) — by writing either of
doIt (Just 2) (25 :: Int)
doIt (Just 2) 25 :: Int -- N.B. this annotates the type of the whole expression
then the specific instance is known.
Note that you do not need type families to produce this behavior. This is par for the course in typeclass resolution. The following code will produce the same error for the same reason.
class Foo a where
foo :: a -> a
main = print $ foo 42
You might be wondering why this doesn't happen with something like
main = print 42
which is a good question, that leftroundabout has already addressed. It has to do with Haskell's defaulting rules, which are so specialized that I consider them little more than a hack.

Answer 3

With this expression:
putStrLn $ show $ tryMe (Just 2) 25
We've got this starting information to work from:
putStrLn :: String -> IO ()
show :: Show a => a -> String
tryMe :: Maybe Int -> Int -> Int
Just :: b -> Maybe b
2 :: Num c => c
25 :: Num d => d
(where I've used different type variables everywhere, so we can more easily consider them all at once in the same scope)
The job of the type-checker is basically to find types to choose for all of those variables, so and then make sure that the argument and result types line up, and that all the required type class instances exist.
Here we can see that tryMe applied to two arguments is going to be an Int, so a (used as input to show) must be Int. That requires that there is a Show Int instance; indeed there is, so we're done with a.
Similarly tryMe wants a Maybe Int where we have the result of applying Just. So b must be Int, and our use of Just is Int -> Maybe Int.
Just was applied to 2 :: Num c => c. We've decided it must be applied to an Int, so c must be Int. We can do that if we have Num Int, and we do, so c is dealt with.
That leaves 25 :: Num d => d. It's used as the second argument to tryMe, which is expecting an Int, so d must be Int (again discharging the Num constraint).
Then we just have to make sure all the argument and result types line up, which is pretty obvious. This is mostly rehashing the above since we made them line up by choosing the only possible value of the type variables, so I won't get into it in detail.
Now, what's different about this?
putStrLn $ show $ doIt (Just 2) 25
Well, lets look at all the pieces again:
putStrLn :: String -> IO ()
show :: Show a => a -> String
doIt :: Test t => TT t -> t -> t
Just :: b -> Maybe b
2 :: Num c => c
25 :: Num d => d
The input to show is the result of applying doIt to two arguments, so it is t. So we know that a and t are the same type, which means we need Show t, but we don't know what t is yet so we'll have to come back to that.
The result of applying Just is used where we want TT t. So we know that Maybe b must be TT t, and therefore Just :: _b -> TT t. I've written _b using GHC's partial type signature syntax, because this _b is not like the b we had before. When we had Just :: b -> Maybe b we could pick any type we liked for b and Just could have that type. But now we need some specific but unknown type _b such that TT t is Maybe _b. We don't have enough information to know what that type is yet, because without knowing t we don't know which instance's definition of TT t we're using.
The argument of Just is 2 :: Num c => c. So we can tell that c must also be _b, and this also means we're going to need a Num _b instance. But since we don't know what _b is yet we can't check whether there's a Num instance for it. We'll come back to it later.
And finally the 25 :: Num d => d is used where doIt wants a t. Okay, so d is also t, and we need a Num t instance. Again, we still don't know what t is, so we can't check this.
So all up, we've figured out this:
putStrLn :: String -> IO ()
show :: t -> String
doIt :: TT t -> t -> t
Just :: _b -> TT t
2 :: _b
25 :: t
And have also these constraints waiting to be solved:
Test t, Num t, Num _b, Show t, (Maybe _b) ~ (TT t)
(If you haven't seen it before, ~ is how we write a constraint that two type expressions must be the same thing)
And we're stuck. There's nothing further we can figure out here, so GHC is going to report a type error. The particular error message you quoted is complaining that we can't tell that TT t and Maybe _b are the same (it calls the type variables a0 and a1), since we didn't have enough information to select concrete types for them (they are ambiguous).
If we add some extra type signatures for parts of the expression, we can go further. Adding 25 :: Int1 immediately lets us read off that t is Int. Now we can get somewhere! Lets patch that into the constrints we had yet to solve:
Test Int, Num Int, Num _b, Show Int, (Maybe _b) ~ (TT Int)
Num Int and Show Int are obvious and built in. We've got Test Int too, and that gives us the definition TT Int = Maybe Int. So (Maybe _b) ~ (Maybe Int), and therefore _b is Int too, which also allows us to discharge that Num _b constraint (it's Num Int again). And again, it's easy now to verify all the argument and result types match up, since we've filled in all the type variables to concrete types.
But why didn't your other attempt work? Lets go back to as far as we could get with no additional type annotation:
putStrLn :: String -> IO ()
show :: t -> String
doIt :: TT t -> t -> t
Just :: _b -> TT t
2 :: _b
25 :: t
Also needing to solve these constraints:
Test t, Num t, Num _b, Show t, (Maybe _b) ~ (TT t)
Then add Just 2 :: Maybe Int. Since we know that's also Maybe _b and also TT t, this tells us that _b is Int. We also now know we're looking for a Test instance that gives us TT t = Maybe Int. But that doesn't actually determine what t is! It's possible that there could also be:
instance Test Double where
type TT Double = Maybe Int
doIt (Just a) _ = fromIntegral a
doIt Nothing b = b
Now it would be valid to choose t as either Int or Double; either would work fine with your code (since the 25 could also be a Double), but would print different things!
It's tempting to complain that because there's only one instance for t where TT t = Maybe Int that we should choose that one. But the instance selection logic is defined not to guess this way. If you're in a situation where it's possible that another matching instance should exist, but isn't there due to an error in the code (forgot to import the module where it's defined, for example), then it doesn't commit to the only matching instance it can see. It only chooses an instance when it knows no other instance could possibly apply.2
So the "there's only one instance where TT t = Maybe Int" argument doesn't let GHC work backward to settle that t could be Int.
And in general with type families you can only "work forwards"; if you know the type you're applying a type family to you can tell from that what the resulting type should be, but if you know the resulting type this doesn't identify the input type(s). This is often surprising, since ordinary type constructors do let us "work backwards" this way; we used this above to conclude from Maybe _b = Maybe Int that _b = Int. This only works because with new data declarations, applying the type constructor always preserves the argument type in the resulting type (e.g. when we apply Maybe to Int, the resulting type is Maybe Int). The same logic doesn't work with type families, because there could be multiple type family instances mapping to the same type, and even when there isn't there is no requirement that there's an identifiable pattern connecting something in the resulting type to the input type (I could have type TT Char = Maybe (Int -> Double, Bool).
So you'll often find that when you need to add a type annotation, you'll often find that adding one in a place whose type is the result of a type family doesn't work, and you'll need to pin down the input to the type family instead (or something else that is required to be the same type as it).
1 Note that the line you quoted as working in your question main = putStrLn $ show $ doIt (Just 2) 25::Int does not actually work. The :: Int signature binds "as far out as possible", so you're actually claiming that the entire expression putStrLn $ show $ doIt (Just 2) 25 is of type Int, when it must be of type IO (). I'm assuming when you really checked it you put brackets around 25 :: Int, so putStrLn $ show $ doIt (Just 2) (25 :: Int).
2 There are specific rules about what GHC considers "certain knowledge" that there could not possibly be any other matching instances. I won't get into them in detail, but basically when you have instance Constraints a => SomeClass (T a), it has to be able to unambiguously pick an instance only by considering the SomeClass (T a) bit; it can't look at the constraints left of the => arrow.