I am playing with Control.Applicative
and I am realizing I don't understand everything with the Haskell type system.
Here is my experiment in Ghci:
λ :t (<*>)
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
λ :t (<*>) (pure 2)
(<*>) (pure 2) :: (Num (a -> b), Applicative f) => f a -> f b
The type of the first argument of <*>
is f (a -> b)
.
- Why is this expression correct?
- How can it be unified with
(pure 2)
since the constant2
is not of typea -> b
? - What does
Num (a -> b)
mean? How can a function having aa -> b
type be an instance ofNum
?
The first argument of
<*>
is supposed to bef (a -> b)
. So given(<*>) (pure x)
, this is well-typed provided thatx
is some kind of function.The type of
2
isNum a => a
. In other words,2
can be any possible type, so long as it's an instance ofNum
.So in your expression
(<*>) (pure 2)
, this is well-typed provided that the type of2
is a function type, and that function type has aNum
instance.Of course, there is almost no reason why you would ever want a function to have a
Num
instance. But the compiler doesn't know that. All it's saying is that if there was such an instance, then the expression would become well-typed.(This is similar to the error you sometimes see where the compiler wants some type to be an instance of
Integral
andFractional
simultaneously. To a human, this is a nonsensical combination. To a machine, they're just two ordinary classes...)