[Haskell-cafe] Type question re: map
Luke Palmer
lrpalmer at gmail.com
Fri Mar 6 23:35:19 EST 2009
2009/3/6 R J <rj248842 at hotmail.com>
>
> Given the following (usual) definition of "map":
>
> map :: (a -> b) -> [a] -> [b]
>
> What's the type of "map map"?
>
The definition is irrelevant, so I removed it.
To make it easier to reason about, I'm going to rename the second map to
map'. It means the same thing, but this is just so we can talk about each
"instance" of it clearly. Now I'm going to rename the free variables:
map :: (a -> b) -> [a] -> [b]
map' :: (c -> d) -> [c] -> [d]
-> is right-associative, so I'll add the implied parentheses:
map :: (a -> b) -> ([a] -> [b])
map' :: (c -> d) -> ([c] -> [d])
Whenever we have an application like f x, if f has type a -> b and x has
type a, then f x has type b.
So map map' says that (a -> b) should be unified with the type of map', and
then the type of the result will be ([a] -> [b]). So the equation is:
a -> b = (c -> d) -> ([c] -> [d])
Which implies
a = c -> d
b = [c] -> [d]
That's as far as we can go with the unification, so the result will be [a]
-> [b]. Substituting, we have [c -> d] -> [[c] -> [d]].
Does that help? What I have done is more-or-less the Hindley-Milner type
inference algorithm.
Luke
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