inferred type of function in Haskell

Question

I defined a function as follows in Haskell

func x | x > 0 = 4
       | otherwise = error "Non positive number"

Its inferred type is

func :: (Ord a, Num a, Num t) => a -> t

Why can't its type be

func :: (Ord a, Num a) => a -> a

Answer 1

The type func :: (Ord a, Num a) => a -> a means that returned type should match with one you passed in. But since you are returning a constant and it doesn't depend on input, its type can be any type of class Num. Thus, compiler infers func :: (Ord a, Num a, Num t) => a -> t.

Answer 2

The inferred type is correct. Let's see how it gets inferred. We will only look at the returned value first:

func x | … = 4
       | … = error "…"

The latter term, error "…" :: a, is unconstrained. However, 4 has the type Num a => a, therefore your function will have type

func :: (Num result) => ????? -> result

Now let's have a look at x:

func x | x > 0 = …
       | …   

Due to the use of (>) :: Ord a => a -> a- > Bool and 0 :: Num a, x's has to be an instance of both Num and Ord. So from that point of perspective, we have

func :: (Num a, Ord a) => a -> ????

The important part is that 4 and x don't interact. Therefore func has

func :: (Num a, Ord a, Num result) => a -> result

Of course, result and a could be the same type. By the way, as soon as you connect x and 4, the type will get more specific:

func x | x > 0 = 4 + x - x
       | …

will have type (Num a, Ord a) => a -> a.