There is no magic here. This is merely explicit type specialization from the most general inferred type to something more specific. The denotational semantics of a function whose type is specialized does not change for those values belonging to the more specialized type.

f :: forall a. (Num a) => a -> a -> a
f x y = x + y

g :: Int -> Int -> Int
g x y = x + y

f and g denote (extensionally) the identical function, differing only in their type. g is a specialization of f. It is possible that (operationally) f could be slower if the compiler is not clever enough to avoid passing a type witness dictionary.

h :: forall b. b -> Char
h = const 'a'

k :: () -> Char
k = const 'a'

data Void
m :: Void -> Char
m = const 'a'

n :: (forall a. a) -> Char
n = const 'a'

h, k, m, and n yield *identical* values for any input compatible with the type of any two of the functions.

In constrast, whether a function is strict or non-strict is *not* a function of type specialization. Strictness is not reflected in the type system. Compare:

u x y = y `seq` const x y
v x y = const x y

Both have type forall a b. a -> b -> a but are denotationally different functions:

u 2 undefined = undefined
v 2 undefined = 2

Cristiano Paris wrote:
On Wed, Sep 16, 2009 at 7:12 PM, Ryan Ingram <ryani.s...@gmail.com> wrote:
Here's the difference between these two types:

test1 :: forall a. a -> Int
-- The caller of test1 determines the type for test1
test2 :: (forall a. a) -> Int
-- The internals of test2 determines what type, or types, to instantiate the
argument at

I can easily understand your explanation for test2: the type var a is
closed under existential (?) quantification. I can't do the same for
test1, even if it seems that a is closed under universal (?)
quantification as well.

Or, to put it another way, since there are no non-bottom objects of type
(forall a. a):

Why?

test1 converts *anything* to an Int.

Is the only possible implementation of test1 the one ignoring its
argument (apart from bottom of course)?

test2 converts *nothing* to an Int

-- type-correct implementation
-- instantiates the (forall a. a) argument at Int
test2 x = x

-- type error, the caller chose "a" and it is not necessarily Int
-- test1 x = x

-- type-correct implementation: ignore the argument
test1 _ = 1

Cristiano
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